Mathematical Analysis a straightforward approach K.G. BIN MORE This is an unf ussy account of mathematical analysis for first year courses at universities and polytechnics that places little reliance on the reader's previous mathematical background. The concepts involved are profound and subtle and their application in particular instances requires a mastery of technique which comes only with time and experience. It is tempting to introduce the concepts at a relatively high level of abstraction; it then becomes possible to expose the elegance of the underlying theory. But such treatments make great demands on a student's technical facility. At the other extreme is the danger of overwhelming students with technical detail. Professor Binmore's approach is to confine attention to 'bread and butter' analysis on the real line and, within this framework, to discuss only those technicalities which are immediately relevant to the basic concepts, the object being to promote confidence and understanding rather than awe or mindless computational efficiency. Emphasis is placed on the importance of tackling problems and the exercises are intended as an integral part of the text. Their worked solutions, which may be consulted when difficulty arises in solving a problem, occupy approximately a third of the book. This abundance of exercises and solutions makes the text highly suitable for those who want to learn the subject on their own as well as for those attending formal classes. C.U.P. NET IN UX £ 9-75 PRESTON POLYTECHNIC LIBRARY & LEARNING RESOURCES SERVICE This book rrust be returned on or before the date last stamped ■■i nc i^~*^ 9 * 31. SEP. 198! <5X- >A/ . -1 HK 1996 p\n 515 BIN PLI Chorley Lancaste Poulton Preston 30107 A/C 095640 000 5: 5 408 :J£43 3T dr^ JpL, MCM tcvn ^y . II MATHEMATICAL ANALYSIS A STRAIGHTFORWARD APPROACH K.G.BINMORE Professor of Mathematics L.S.E. CAMBRIDGE UNIVERSITY PRESS CAMBRIDGE LONDON - NEW YORK - MELBOURNE Published by the Syndics of the Cambridge University Press The Pitt Building, Trumpington Street, Cambridge CB2 1 RP Bentley House, 200 Euston Road, London NW1 2DB 32 East 57th Street, New York, NY 10022, USA 296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia © Cambridge University Press 1977 First published 1977 Printed in Great Britain at the University Press, Cambridge Library of Congress Cataloguing in Publication Data Binmore,KG 1940- Mathemalical analysis. Includes index. 1. Mathematical analysis. I. Title. QA300.B536 515 76-28006 ISBN 521 21480 7 hard covers ISBN 521 29167 4 paperback CONTENTS Preface IX 1 Real numbers 1 1.1 Set notation 1 1 2 The set of real numbers 2 1 3 Arithmetic 3 1 4 Inequalities 3 1 9 Roots 6 1 10 Quadratic equations 6 1 13- Irrational numbers 9 1 14 Modulus 10 2 Continuum Property 12 2.1 Achilles and the tortoise 12 2.2 The continuum property 13 2.6 Supremum and infimum 15 2.7 Maximum and minimum 15 2.9 Intervals ■:•> 16 2.11 Manipulations with sup and inf 18 3 Natural numbers 20 3.1 Introduction 20 3.2 Archimedean property 20 3.7 Principle of induction 22 4 Convergent sequences 26 4.1 The bulldozers and the bee 26 4.2 Sequences 27 4.4 Definition of convergence 27 4.7 Criteria for convergence 30 4.15 Monotone sequences 33 4.21 Some simple properties of convergent sequences 37 4.26 Divergent sequences 38 5 Subsequences 41 5.1 Subsequences 41 5.8 Bolzano-Weierstrass theorem 46 VI Contents ™ * Contents vii 5.12 Lim sup and lim inf 47 12 Monotone functions 108 5.16 Cauchy sequences 49 12.1 Definitions 108 12.3 Limits of monotone functions 108 6 Series 53 12.6 Differentiable monotone functions 110 6.1 Definitions 53 12.9 Inverse functions 110 6.4 Series of positive terms 54 12.11 Roots 112 6.7 Elementary properties of series 56 12.13 Convex functions 114 6.12 Series and Cauchy sequences 57 13 Integration 6.20 Absolute and conditional convergence 60 119 6.23 Manipulations with series 61 13.1 Area 119 13.2 The integral 120 7 Functions 64 13.3 Some properties of the integral 121 7.1 Notation 64 13.9 Differentiation and integration 124 7.6 Polynomial and rational functions 66 13.16 Riemann integral 127 7.9 Combining functions 68 13.19 More properties of the integral 129 7.11 Inverse functions 68 13.27 Improper integrals 132 7.13 Bounded functions 70 13.31 Euler-Maclaurin summation formula 134 8 Limits of functions 74 14 Exponential and logarithm 137 14.1 - Logarithm 137 8.1 Limits from the left 74 14.4 Exponential 140 8.2 Limits from the right 74 14.6 Powers 141 8.3 /(x)^/as:c-*£ 75 8.6 Continuity at a point 77 15 Power series 143 8.8 Connexion with convergent sequences 78 15.1 Interval of convergence 143 8.11 Properties of limits 79 15.4 Taylor series 145 8.16 8.18 Limits of composite functions Divergence 81 82 15.7 Continuity and differentiation 147 16 Trigonometric functions 150 9 Continuity 84 16.1 Introduction 150 9.1 Continuity on an interval 84 16.2 Sine and cosine 151 9.7 Continuity property 86 16.4 Periodicity 153 10 Differentiation 91 17 The gamma function 156 10.1 Derivatives 91 17.1 Stirling's formula 156 10.2 Higher derivatives 92 17.3 The gamma function 157 10.4 More notation 93 17.5 Properties of the gamma function 158 10.5 Properties of differentiable functions 95 18 10.12 Composite functions 98 Appendix 161 This contains the proofs of 'propositions' left unproved in the main 11 Mean value theorems 100 . body of the text. 11.1 Local maxima and minima 100 Solutions to exercises 172 11.3 Stationary points 101 11.5 Mean value theorem 102 Suggested further reading 251 11.9 Taylor's theorem 105 Notation Index 253 255 PREFACE This book is intended as an easy and unfussy introduction to mathematical analysis. Little formal reliance is made on the reader's previous mathematical background, but those with no training at all in the elementary techniques of calculus would do better to turn to some other book. An effort has been made to lay bare the bones of the theory by eliminating as much unnecessary detail as is feasible. To achieve this end and to ensure that all results can be readily illustrated with concrete examples, the book deals only with 'bread and butter' analysis on the real line, the temptation to discuss gener- alisations in more abstract spaces having been reluctantly suppressed. However, the need to prepare the way for these generalisations has been kept well in mind. It is vital to adopt a systematic approach when studying mathematical analy- sis. In particular, one should always be aware at any stage of what may be assumed and what has to be proved. Otherwise confusion is inevitable. For this reason, the early chapters go rather slowly and contain a considerable amount of material with which many readers may already be familiar. To neglect these chapters would, however, be unwise. The exercises should be regarded as an integral part of the book. There is a great deal more to be learned from attempting the exercises than can be obtained from a passive reading of the text. This is particularly the case when, as may fre- quently happen, the attempt to solve a problem is unsuccessful and it is necess- ary to turn to the solutions provided at the end of the book. To help those with insufficient time at their disposal to attempt all the exer- cises, the less vital exercises have been marked with the symbol t- (The same notation has been used to mark one or two passages in the text which can be omitted without great loss at a first reading.) The symbol * has been used to mark exercises which are more demanding than most but which are well worth attempting. The final few chapters contain very little theory compared with the number of exercises set. These exercises are intended to illustrate the power of the tech- niques introduced earlier in the book and to provide the opportunity of some re- vision of these ideas. This book arises from a course of lectures in analysis which is given at the London School of Economics. The students who attend this course are mostly not specialist mathematicians and there is little uniformity in their previous IX Preface x mathematical training. They are, however, quite well-motivated. The course is a 'one unit' course of approximately forty lectures supplemented by twenty in- formal problem classes. I have found it possible to cover the material of this book in some thirty lectures. Time is then left for some discussion of point set topology in simple spaces. The content of the book provides an ample source of examples for this purpose while the more general theorems serve as reinforce- ment for the theorems of the text. Other teachers may prefer to go through the material of the book at a more leisurely pace or else to move on to a different topic. An obvious candidate for further discussion is the algebraic foundation of the real number system and the proof of the Continuum Property. Other alternatives are partial differentiation, the complex number system or even Lebesgue measure on the line. I would like to express my gratitude to Elizabeth Boardman and Richard Holmes for reading the text for me so carefully. My thanks are also due to 'Buffy' Fennelly for her patience and accuracy in preparing the typescript. Final- ly, I would like to mention M.C. Austin and H. Kestelman from whom I learned so much of what I know. July 1976 K.G.B. 1 REAL NUMBERS 1.1 Set notation A set is a collection of objects which are called its elements. If x is an element of the set S, we say that x belongs to S and write xes. Ify does not belong to 5, we write y £ S. The simplest way of specifying a set is by listing its elements. We use the notation A = &1, >£,«,*} to denote the set whose elements are the real numbers \, 1 , \Jl, e and it. Similarly B = {Romeo, Juliet} denotes the set whose elements are Romeo and Juliet. This notation is, of course, no use in specifying a set which has an infinite number of elements. Such sets may be specified by naming the property which distinguishes elements of the set from objects which are not in the set. For example, the notation C = {x : x > 0} (which should be read 'the set of all x such that x > 0') denotes the set of all positive real numbers. Similarly D = {y : y loves Romeo} denotes the set of all people who love Romeo. It is convenient to have a notation for the empty set 0. This is the set which has no elements. For example, if x denotes a variable which ranges over the set of all real numbers, then {x : x 2 + 1 = 0} = 0. This is because there are no real numbers x such that x 2 = — 1 . If 5 and T are two sets, we say that 5 is a subset of T and write Real numbers SCT if every element of S is also an element of T. As an example, consider the sets P = {1 , 2, 3 , 4} and Q = {2, 4}. Then QCP. Note that this is not the same thing as writing Q e P, which means that Q is an element of P. The elements ofP are simply 1,2,3 and 4. But Q is not one of these. The sets A , B, C and D given above also provide some examples. We have A C Cand (presumably) BCD. 1 .2 The set of real numbers It will be adequate for these notes to think of the real numbers as being points along a straight line which extends indefinitely in both directions. The line may then be regarded as an ideal ruler with which we may measure the lengths of line segments in Euclidean geometry. n/2 -2 The set of all real numbers will be denoted by R . The table below distin- guishes three important subsets of U. Subset Notation Elements Natural numbers (or whole numbers) Integers Rational numbers (or fractions) Q 1,2,3,4,5,. .. . .-2,-1,0, 1,2,3, 0,1,2.-1.*, If. -I, -I Not all real numbers are rational. Some examples of irrational numbers are v2, e and ir. While we do not go back to first principles in these notes, the treatment will be rigorous in so far as it goes. It is therefore important to be clear, at every stage, about what our assumptions are. We shall then know what has to be proved and what may be taken for granted. Our most vital assumptions are con- cerned with the properties of the real number system. The rest of this chapter and the following two chapters are consequently devoted to a description of the Real numbers properties of the real number system which we propose to assume and to some of their immediate consequences. 1.3 Arithmetic The first assumption is that the real numbers satisfy all the usual laws of addition, subtraction, multiplication and division. The rules of arithmetic, of course, include the proviso that division by zero is not allowed. Thus, for example, the expression 2 makes no sense at all. In particular, it is not true that We shall have a great deal of use for the symbol °°, but it must clearly be understood that °° does not represent a real number. Nor can it be treated as such except in very special circumstances. 1 .4 Inequalities The next assumptions concern inequalities between real numbers and their manipulation. We assume that, given any two real numbers a and b, there are three mutually exclusive possibilities: (i) a > b (a is greater than b) (ii) a = b (a equals b) (iii) a < b (a is less than b). Observe that a < b means the same thing as b > a. We have, for example, the following inequalities. 1 >0; 3>2; 2<3; -K0; -3<-2. There is often some confusion about the statements (iv) a~>b (a is greater than or equal to b) (v) a < b (a is less than or equal to b). To clear up this confusion, we note that the following are all true statements. 1 >0\ 3>2; 1 > 1; 2<3; -1 <0; -3«— 3. We assume four basic rules for the manipulation of inequalities. From these 4 Real numbers the other rules may be deduced. (I) If a >b and b >c, then a >c. (II) If a > b and c is any real number, then a+c>b+c. (III) If a > b and c > 0, then ac > 6c (i.e. inequalities can be multiplied through by a positive factor). (IV) If a > b and c < 0, then ac < be (i.e. multiplication by a negative factor reverses the inequality). 7.5 Example If a > 0, prove that a -1 > 0. /Voo/ We argue by contradiction. Suppose that a > but that a -1 < 0. It cannot be true that a" 1 = (since then = O.a = 1). Hence a _1 <0. By rule III we can multiply this inequality through by a (since a > 0). Hence 1 = a"'.a<0.a = 0. But 1 < is a contradiction. Therefore the assumption a -1 < was false. Hence a _I >0. 1.6 Example If x and y are positive, thenar <y if and only if x 2 <y 2 . Proof We have to show two things. First, that x < y implies x 2 < y 2 , and secondly, that x 2 < y 2 implies x < y. (i) We begin by assuming that* <y and try to deduce that x 2 <y 2 . Multiply the inequality x < y through by x > (rule III). We obtain x 2 <xy. Similarly xy<y 2 . But now* 2 <y 2 follows from rule I. (ii) We now assume that x 2 <y 2 and try and deduce that* < y. Adding —x 2 to both sides of* 2 < y 2 (rule II), we obtain y 2 -x 2 >0 i.e. (y~x)(y + x)>0. (1) Real numbers • Since x + y > 0, (x +y)~ 1 > (example 1 .6). We can therefore multiply through inequality (1) by (x + y)' 1 to obtain y-x>0 i.e. x<y. (Alternatively, we could prove (ii) as follows. Assume that* 2 <y 2 but that x > y. From x > y it follows (as in (i)) that x 2 > y 2 , which is a contradiction.) 1.7 Example Suppose that, for any e>0, a <b + e. Thena <fc. Proof Assume that a > b. Then a — b > 0. But, for any e > 0, a < b + e. Hence a < b + e in the particular case when e = b —a. Thus a<b + (a-b) and so a<a. This is a contradiction. Hence our assumption a > b must be false. Therefore a<b. (Note: The symbol e in this example is the Greek letter epsilon. It should be carefully distinguished from the 'belongs to' symbol G and also from the symbol % which is the Greek letter*/'.) 1.8 Exercise (1) If* is any real number, prove that* 2 >Q. If 0<a< 1 andb > 1, prove that (i)0<a 2 <a<l (ii) b 2 >b> I. (2) IfZ>>0and£>0and b B' prove that ali < bA . Deduce that a a + A A ~b < b + B B' (3) If a > ft and c > d, prove that a +c>b+d(i.e. inequalities can be added). If, also, b > and d > 0, prove that ac > bd (i.e. inequalities between positive numbers can be multiplied). (4) Show that each of the following inequalities may fail to hold even though a>Z>andc>d. (i)a-c>b-d Real numbers <>1 (iii) ac>bd. What happens if we impose the extra condition that b > and d > 0? (5) Suppose that, for any e>0,a — e<b<a + e. Prove that a = b . (6) Suppose that a < b. Show that there exists a real number x satisfying a<x<b. 1.9 Roots Let n be a natural number. The reader will be familiar with the notation y =x". For example, x 2 =x.x and* 3 = x.x.x. Our next assumption about the real number system is the following. Given any y>0 there is exactly one value of x > such that y = x". (Later on we shall see how this property may be deduced from the theory of continuous functions.) If y > 0, the value of x > which satisfies the equation y = x n is called the nth root of y and is denoted by x = y Un . When n = 2, we also use the notation y/y = y xn - Note that, with this con- vention, it is always true that y/y > 0. If y > 0, there are, of course, two num- bers whose square is y. The positive one is y/y and the negative one fc—y/y. The notation ± y/y means 'y/y or —y/y'. If r = m/n is a positive rational number and y > 0, we define y r = (y m y ,n . If r is a negative rational, then — r is a positive rational and hence y~ r is defined. If y > we can therefore define y r by Y = 1 We also write y° = 1 . With these conventions it follows that, if y > 0, then ^ r is defined for all rational numbers r. (The definition of y x when x is an irrational real number must wait until a later chapter.) 1.10 Quadratic equations If j> > 0, the equation x 2 =y has two solutions. We denote the positive solution by y/y. The negative solution is therefore —y/y. We note again that Real numbers there is no ambiguity about these symbols and that ± y/y simply means 'y/y or -y/y. The genera] quadratic equation has the form ax 2 + bx + c = where a ¥= 0. Multiply through by 4a. We obtain 4a 2 x 2 + Aabx + 4ac = {lax +b) 2 -b 2 +4ac = (2ax + b) 2 = b 2 -4ac. It follows that the quadratic equation has no real solutions if b 2 — 4ac < 0, one real solution if b 1 —4ac = and two real solutions if A 2 — 4ac > 0. If b 2 - 4ac > 0, 2ax+b = ±y/(b 2 -4ac) -b ±y/(b 2 -4ac) x = 2a The roots of the equation ax 2 + bx + c = are therefore -b-y/(b 2 -4ac) , . - b + y/(b 2 - 4ac) a = and u = ~ . 2a v 2a It is a simple matter.to check that, for all values of x, ax 2 +bx + c = a(x-a)(x-p). With the help of this formula, we can sketch the graph of the equation y -ax 2 + bx + c. ' y = ax 2 + bx +c V fl>0 y = ax J - bx + e 7^ a<0 1.11 Example A nice application of the work on quadratic equations described above is the proof of the important Cauchy Schwarz inequality. This asserts that, if a,, a 2 , . . . ,a„ andZ>i,Z» 2 , . . . ,fe„ are any real numbers, then (a,6, +a 2 b? Bnh? < («i a\ ... + a 2 n ){b 2 +b\ b 2 n ). 8 Real numbers Proof For any x, < (a l x + b l ) 2 + (a 2 x+b 2 ) 2 + ... + (a n x+b n ) 2 = (a? + . . . + a 2 n )x 2 + 2ia.br + ... + a n h n )x + (b\ + . . . + b 2 n ) = Ax 2 + 2Bx + C. Since y = Ax 2 + 2Bx + O for all values of x, it follows that the equation Ax 2 + 2Bx + C = cannot have two (distinct) roots. Hence (2B) 2 -4AC<0 i.e. B 2 <AC which is what we had to prove. 1.12 Exercise (1) Suppose that n is an even natural number. Prove that the equation a-" = y has no solutions if y < 0, one solution if y = and two solutions if y > 0. Suppose that n is an odd natural number. Prove that the equation x" = y always has one and only one solution. Draw graphs of y = x 2 and y = x 3 to illustrate these results. (2) Simplify the following expressions: (OS 2 ' 3 (ii)27" 4 ' 3 (iii)32 6/s . (3) If y > 0, z > and r and s are any rational numbers, prove the follow- ing: (■) y r+s = ff (ii) y rs = (yj (iii) (yzj = fz\ (4) Suppose that a > and that a and are the roots of the quadratic equation ax 2 + bx + c = (in which b 2 — 4ac > 0). Prove that y = ax 2 + bx + c is negative when a < x < and positive when x < a orx >j3. Show also (without the use of calculus) that y = ax 2 + bx + c achieves a minimum value of c —b 2 /4a when* = —bj2a. (5) Leta,,a 2 , . . . , a„ be positive real numbers. Their arithmetic meanA n and harmonic mean //„ are defined by _ g,+g 2 + ...+a„ III. 1 H-' = -- + -+... + - n\ai a 2 a„ Deduce from the Cauchy-Schwarz inequality that H n < A„. (6) Letai,a 2) ... ,a n andb i ,b 2 , ■ ■ ■ ,b n beany real numbers. Prove Minkowski's inequality, i.e. I (^ + b k y 1/2 < n 1/2 n >,al + *:# k = i «t=i /?ea/ numbers i For the case « = 2 (or « = 3) this inequality amounts to the assertion that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. Explain this. 1.13 Irrational numbers In § 1 .2 we mentioned the existence of irrational real numbers. That such numbers exist is by no means obvious. For example, one may imagine the process of marking all the rational numbers on a straight line. First one would mark the integers. Then one would move on to the multiples of j and then to the multiples of i and so on. Assuming that this program could ever be com- pleted, one might very well be forgiven for supposing that there would be no room left for any more points on the line. h tt i * iUtlLM -1 But our assumption about the existence of nth roots renders this view unten- able. This assumption requires us to accept the existence of a positive real num- ber* (namely \J2) which satisfies x 2 = 2. If* were a rational number it would be expressible in the form m x = — n where m and n are natural numbers with no common divisor (other than 1). It follows that 2 = In 2 m and so m 2 is even. This implies that m is even. (If m were odd, we should have m = 2k + 1 . But then m 2 = 4k 2 + 4& + 1 which is odd.) We may therefore write m = 2k. Hence 4k 2 = 2n 2 n 2 = 2k 2 . Thus n is even. We have therefore shown that both m and n are divisible by 2. This is a contradiction and it follows that x cannot be rational, i.e. \/2 must be an irrational real number. Of course, \/2 is not the only irrational number and the ability to extract «th roots allows us to construct many others. But it should not be supposed that all irrational numbers can be obtained in this way. It is not even true that every irrational number is a root of an equation of the form 10 Real numbers a + a x x + a 2 x 2 + . . . + a n x" = in which the coefficients a ,a t , ■ ■ ■ ,o n ate rational numbers. Real numbers which are not the roots of such an equation are called transcendental, notable examples being e and it. This fascinating topic, however, lies outside the scope of this book. 1.14 Modulus Suppose that x is a real number. Its modulus (or absolute value) \x\ is defined by Ay 1*1 = x (x>0) -x (x < 0). Thus |3 1 = 3, 1—41 = 6 and |0| = 0. Obviously |*| > for all values of*. It is sometimes useful to note that | x | = s/x 2 . 1.15 Theorem For any real numbers:, -|x| <x<|x|. Proof Either x > or x < 0. In the first case, — |x| < < * = |x|. In the second case, — 1*| — X <0 < |*|. 1.16 Tlieorem For any real numbers a and b \ab\ = \a\.\b\ Proof The most elegant proof is the following. \ab\ = s/((abf) = y/(a 2 b 2 ) = sj(a 2 ).yj(b 2 ) = \a\.\b\. The next theorem is of great importance and is usually called the triangle inequality. This nomenclature may be justified by observing that the theorem is the special case of Minkowski's inequality (exercise 1 .12(6)) with n = 1 . We give a separate proof. 1.1 7 Theorem (triangle inequality ) For any real numbers a and b , \a + b\ < \a\+ |6|. Proof We have Real numbers 11 \a+b\ 2 = {a + bf = a 2 + lab + b 2 = \a\ 2 + 2ab + \b\ 2 < \a\ 2 + 2\ab\+ \b\ 2 (theorem 1.15) = \a\ 2 + 2\a\-\b\ + \b\ 2 (theorem 1.16) = (\a\+\b\) 2 . It follows (from example 1 .6) that \a + b\ < |o|+ \b\. 1.18 Theorem For any real numbers c and d, \c-d\ > |c|-|d|. Proof Take a = d and b = c — d in the triangle inequality. Then a + b =c and so |c| = \a + b\ < |a|+ \b\ = \d\ + \c-d\. 1.19 Example (i) 2 = |-1+3|<|-1|+|3| = 1 +3=4.' (ii)7 = |6-(-l)|> |6|-|-1| = 6-1 =5. 1.20 Exercise (1) Prove that |a| <b if and only if —b <a<b. [Recall that you have two things to prove as in example 1 .6.] (2) Prove that \c-d\ > \\c\-\d\\. (3) The distance d(x,y) between two real numbers x undy is defined by d(x,y) = |* —y\. Show that, for any x,y andz, (i) d(x,y)>0 (iii)d(;c,.y) = 0ifand only if x=y (ii) d(x,y)=d(y,x) (iv)d(x,y)<d(x,z) + d(z,y). 1 (4) If r and s are rational numbers, prove that r + sy/2 is irrational unless^ = 0. > (5) Suppose that the coefficients a¥=0,b and c of the quadratic equation ax 2 + bx+c = are all rational numbers and that a = r + s\/2 is a root of this equation, where r and s are also rational numbers. Prove that (3 = r —s\/2 is also a root, t (6) Prove that 3" 2 and 2 1/3 are irrational numbers. CONTINUUM PROPERTY 2.1 Achilles and the tortoise The following is one of the famous paradoxes of Zeno. Achilles is to race a tortoise. Since Achilles runs faster than the tortoise, the tortoise is given a start of x feet. When Achilles reaches the point where the tortoise started, the tortoise will have advanced a bit, say x 1 feet. Achilles soon reaches the tortoise's new position, but, by then, the tortoise will have advanced a little bit more, say x 2 feet. This argument may be continued indefinitely and so Achilles can never catch the tortoise. A ** -* v„ >• A m* ■< x, >■ A - The simplest way to resolve this paradox is to say that Achilles catches the tortoise after he has run a distance of x feet, where x is 'the smallest real number larger than all of the numbers x ,x + x u x + x y x 2 , . . .'. Zeno's argument then simply reduces to subdividing a line segment of length x into an infinite number of smaller line segments of respective lengths x ,x lt x 2 , . . . Formulated in this way, the paradox loses its sting. This solution, of course, depends very strongly on the existence of the real number x, i.e. the smallest real number larger than all of the numbers x , x +*!, jc +x$ +x 2 , . ■ ■ 12 Continuum property 13 A similar lesson is to be learned from Archimedes' method of exhaustion. This was invented by Archimedes as a means of evaluating the area of regions with curved boundaries. In the case of a circle, the method involves inscribing a sequence of larger and larger regular polygons inside the circle and then calculating their areas Ai,A 2 ,A 3 , . . . The area A of the circle is then identified with the 'smallest real number larger than each of the numbers/^, A 2 ,A 3 , . . .'. By taking the radius of the circle to be 1 and calculating the value of A n for a large enough value of n, Archimedes was able to obtain as close an approximation to -n as he chose. Note that this method again depends strongly on the existence of a smallest real number larger than all of the real numbers Ai,A 2 ,A-$,... 2.2 The Continuum Property The discussion of the previous section indicates the necessity of making a further assumption about the real number system. It would, after all, be most unsatisfactory if Achilles never caught the tortoise or if a circle had no area. We shall call our new assumption the Continuum Property of the real numbers. The Continuum Property is not usually mentioned explicitly in a school algebra course, but it is very important for what follows in these notes. The rest of this chapter is therefore devoted to its consideration. We begin with some terminology. A set S of real numbers is bounded above if there exists a real number H which is greater than or equal to every element of the set, i.e. if, for some //, x<H for any x e S. The number H (if such a number exists) is called an upper bound of the set S. A set S of real numbers is bounded below if there exists a real number h 14 Continuum property which is less than or equal to every element of the set, i.e. if, for some h. Continuum property B<H. 15 i B \H x>h for any xGS. The number h (if such a number exists) is called a lower bound of the set S. A set which is both bounded above and bounded below is just said to be bounded. 2.3 Proposition A set S of real numbers is bounded if and only if there exists a real number K such that |jc| < K for any xG S. (This is easily proved with the help of exercise 1 .20(1).) 2.4 Examples (i) The set {1 , 2, 3} is bounded above. Some upper bounds are 100, 10, 4 and 3. The set is also bounded below. Some lower bounds are —27, and 1 . (ii) The set {x : 1 < x < 2} is bounded above. Some upper bounds are 100, 10, 4 and 2. The set is also bounded below. Some lower bounds are —27, and 1. (iii) The set {x : x > 0} is unbounded above. If H > is proposed as an upper bound, one has only to point to H + 1 to obtain an element of the set larger than the supposed upper bound. However, the set {x :x> 0} is bounded below. Some lower bounds are —27 and 0. We now state the Continuum Property which will be fundamental for the remainder of these notes. Continuum Property Every non-empty set of real numbers which is bounded above has a smallest upper bound. Every non-empty set of real numbers which is bounded below has a largest lower bound. Thus, if S is a non-empty set which is bounded above, then S has an upper bound B such that, given any other upper bound //of S, Similarly, if S is a non-empty set which is bounded below, then S has a lower bound b such that, given any other lower bound h, b>h. 2.5 Examples (i) The smallest upper bound of the set {1,2, 3} is 3. The largest lower bound of the set {1,2, 3} is 1. (ii) The smallest upper bound of the set {x : 1 < x < 2} is 2. The largest lower bound is 1 . (iii) The set {x : x > 0} has no upper bounds at all. The largest lower bound of the set {x:x>0}is0. 2.6 Supremum and infimum If a non-empty set S is bounded above, then, by the Continuum Property, it has a smallest upper bound B. This smallest upper bound B is some- times called the supremum of the set S. We write B — sup 5 or B = sup x. i6=S Similarly, a set which is bounded below has a largest lower bound b. We call b the infimum of the set S and write b = inf 5 or b = inf x. xGS Sometimes you may encounter the notation sup S = + °°. This simply means that 5 is unbounded above. Similarly, inf 5 = — °° means that S is unbounded below. 2.7 Maximum and minimum If a set S has a largest element M, we call M the maximum of the set S and write M = max S. If 5 has a smallest element m, we call m the minimum of S 1 and write m = min S. 16 Continuum property It is fairly obvious that, if a set S has a maximum M, then it is bounded above and its smallest upper bound is M. Tlius, in this case, sup S = max S. A common error is to suppose that the smallest upper bound of a set S is always the maximum of the set. However, some sets which are bounded above (and hence have a smallest upper bound) do not have a maximum. (See example 2.8(a).) Similar remarks, of course, apply to largest lower bounds and minima of sets. 2.8 Examples (i) The set {l , 2, 3} has a maximum 3 and this is equal to its smallest upper bound. The set {l , 2, 3} has minimum 1 and this is equal to its largest lower bound. (ii) The set {x : 1 < x < 2} has no maximum. The number 2 cannot be the largest element of the set because it does not belong to the set. On the other hand, any x in the set satisfies 1 < x < 2. But then x + 2 y = is an element of the set which is larger than x. Hence x cannot be the largest element of the set. However, {x : 1 < x < 2} has smallest upper bound 2. The set {x : 1 < x < 2} has minimum 1 (why?). This is equal to its largest lower bound. (iii) The set {x : x > 0} has no maximum, nor does it have any upper bounds. The set {x : x > 0} has no minimum (why not?). Its largest lower bound is 0. 2.9 Intervals An interval I is a set of real numbers with the property that, if x E I and y Eland x<z^y, then zEI, i.e. if two numbers belong to /, then so does every number between them. In describing intervals we use the following notation: (a, b) = {x:a<x<b} [a,b] = {x:a<x<b} [a,b) = {x:a<x<b} (a,b] = {x:a<x<b}. a b wmm (a,.' ) T Continuum property 17 These are the bounded intervals (classified by whether or not they have a maximum and whether or not they have a minimum). We also need to consider the unbounded intervals. For these we use the notation below. (a, °°) = {x : x >a} [a,°°) = {x :x>d) (-°°,b) = {x:x<b} (-°°,b] = {x:x<b} V7/////////'y/7//7/7////y, («,») ■/y////////////7\ W //W////7W (Do not be misled by this notation into supposing that °° or — °° are real num- bers or that they can be treated as such.) All intervals (with the exception of the empty set and the set of all real numbers) fall into one of the categories described above. We call the intervals (a,6),(a,°°)and(— °°,b)open and the intervals [a, b], [a, °°) and (— °°,fe] closed. (The intervals [a, b) and (6, a] are sometimes called half-open.) Of particular importance are the closed, bounded intervals [a, b] . We shall call such an interval compact. 2.10 Exercise (1) Which of the following statements are true? 0)36(1,2) (ii)2e(-°°,3] (iii) 16 [1,2) (iv)2e(0,2) (v)3G[l,3]. (2) If % is a real number and 5 > 0, prove that {x:|£-x|<S} = a-S,£ + 5). (3) Decide in each of the following cases whether or not the given set is bounded above. For those sets which are bounded above, write down three different upper bounds including the smallest upper bound. Decide which of the sets have a maximum and, where this exists, write down its value. (i)(0,l) (ii)(-°°,2] (iii) {-1,0, 2, 5} (iv)(3,°°) (v) L0, 1]. (4) Repeat the above question but with the words 'above, upper and maxi- mum' replaced by 'below, lower and minimum'. (5) Give an example of a set which has smallest upper bound 4 but contains no element x satisfying 3 < x < 4. (6) Show that, given any element x of the set (0, °°), there is another element y 6 (0, °°) with the property that ;' < x. Deduce that (0, °°) has no minimum. 18 Continuum property Continuum property 19 2.11 Manipulations with sup and inf We prove one result and list some others as exercises. 2.12 Theorem Suppose that S is a non-empty set of real numbers which is bounded above and % > 0. Then sup %x = % sup x. Proof Let B = sup S. Then B is the smallest number such that, for any x<ES, x<B. Let 2"- {#* : *€ S}. Since | > 0, for any* G 5. Hence Tis bounded above by £/?. By the Continuum Property, rhas a smallest upper bound (or supremum) C. We have to prove that C = %B. Since %B is an upper bound for 7* and C is the smallest upper bound for 7 1 , Now repeat the argument with the roles of S and T reversed. We know that C is the smallest number such that, for any y&T, y<C. Since £ > 0, it follows that for any y G T. But S = {|"V : ? G 7% Hence ?"' C is an upper bound for S. But # is the smallest upper bound for S. Thus £5<C. We have shown that C< |5 and also that £fi < C. Hence \B = C. 2.13 Exercise (1) Suppose that S is bounded above and that S C S. Prove that sup S < sup 5. (2) Suppose that S is bounded above and that £ is any real number. Prove that sup (x + £) = sup x + %. *£S i£S (3) Suppose that 5 is bounded above. Prove that inf (— x) = — sup x. i£s .*es Hence obtain analogies of theorem 2.12 and exercises 2.13(1 and 2) in which inf replaces sup. (4) The distance d{%, S) between a real number £ and a non-empty set 5 of real numbers is defined by d(S,S) - inf |$-*| i£S (see exercise 1.20(3)). Find the distance between the real number % = 3 and the sets (i)S = {0, 1,2} (ii)5 = (0,l) (iii)S=[l,2] (iv)S = (2,3). (5) (i) If £ G S, prove that d(|, 5) = 0. Give an example of a real num- ber ? and a non-empty set 5 for which d(£, S) = but £ G" S. (ii) If 5 is bounded above and % = sup S, prove that d(|, 5) = 0. Deduce that the same is true if 5 is bounded below and £ = inf S. "I" (iii) If /is a closed interval, prove that d{%, /) = implies that \ G /. If/ is an open interval (other than R or 0), show that a j- ^/can always be found for which d(£, /) = 0. f (6) Suppose that every element of an interval / belongs to one or the other of two non-empty subsets 5 and T. Show that one of the two sets S or T contains an element at zero distance from the other. (Do not assume that S and Tare necessarily intervals. The set S, for example, could consist of all rational numbers in / and the set T of all irrational num- bers in /.) [Hint : Suppose that $ < t, where s G S and t G T. Let T = {x : x G T and x > s}. Show that To ^ and is bounded below. Write b = inf T and consider the two cases b € Tand b ^ T.\ NATURAL NUMBERS 3.1 Introduction In this chapter we explore some implications of the Continuum Property for sets of natural numbers. As always, arithmetical properties are taken for granted. But where a proof hinges on such a property, the property will be explicitly stated. 3.2 Archimedean property The Archimedean property is the assertion of the following theorem. 3.3 Theorem The set M of natural numbers is unbounded above. Proof For this proof we need to assume that, if n is a natural number, then so is n + 1 . Suppose that the theorem is false. Then N is bounded above. By the Con- tinuum Property it therefore has a smallest upper bound B. Since B is the smallest upper bound for N, B — 1 is not an upper bound for N. Thus, for some «€N, n >B-\ n+ I >B. But then n + 1 is an element of the set fsj which is greater than B. But B is an upper bound of the set M. This is a contradiction. Hence M is unbounded above. 3.4 Example Prove that the set 5 = {«-' :neN} is bounded below with largest lower bound 0. Proof Since the natural numbers are all positive, o n _1 >0 20 Natural numbers 21 for all n € foJ. Hence the set S is bounded below by 0. We have to prove that is the largest lower bound . Suppose that /; > is a lower bound for S. Then, for each n € M, n' 1 >h. Hence, for each n e fol , n « ft" 1 . But then h' 1 is an upper bound for the set M which we know is unbounded above. It follows from this contradiction that no h > can be a lower bound for S. Thus is the largest lower bound for S. Note that 0^5 and hence S has no minimum. 3.5 Theorem Every set of natural numbers which is not empty has a mini- mum. Proof For this proof we need to assume that the distance between distinct natural numbers is at least 1 . Let S be a non-empty set of natural numbers. Given that all natural numbers are positive, is a lower bound for S. By the Continuum Property, it follows that S has a largest lower bound b. Since b is the largest lower bound, b + 1 is not a lower bound. Therefore, for some «SS, n < b + 1 . If n is the minimum of S, there is nothing to prove. If not, then, for some wj e S,m< n. We obtain the inequality b < m < n < b + 1 from which follows the contradiction < n — m < 1. 3.6 Exercise (1) Prove that the set n—l s = :«e is bounded above with smallest upper bound 1 . Does S have a maxi- mum? (2) Let X > 1 . Prove that the set S = {X n :n&U} is unbounded above. [Hint: if B were the smallest upper bound, then BX~ l could not be an upper bound.] Show that, if 0<x < 1, then the 22 Natural numbers set T = {x n :nGN} is bounded below with largest lower bound 0. (3) Prove that any non-empty set of integers which is bounded above has a maximum and that any non-empty set of integers which is bounded below has a minimum. (4) Let a < b. Prove that there exists a rational number r satisfying a<r<b. [Hint: justify the existence of a natural number n satisfying n~>(b —a)~ l and consider the rational number r = tn/n where m is the smallest integer satisfying m>an.] (5) Let S be the set of all rational numbers r which satisfy < r < 1 . Show that S has no maximum and no minimum. Prove that S is bounded and has largest lower bound and smallest upper bound 1 . [Hint: for the last part, use the previous question.] (6) Let a < b. Prove that there exists an irrational number £ satisfying a < % < b. [Hint: exercise 1.20(4).] 3.7 Principle of induction Suppose that a line of dominoes is arranged so that, if the «th one falls, it will knock over the (n + l)th. If the first domino is pushed over, most people would agree that all the dominoes would then fall down. The principle of induction is an idealisation of this simple notion. We show below how it may be deduced from theorem 3.5. 3.8 Theorem (principle of induction) Suppose that, for each n £ M , P(n) is a statement about the natural number n. Suppose also that (i)P(l)is true (ii) if P(n) is true, then P(n + 1) is true. Then P(n) is true for every n £ftj . Proof For this proof we need to assume that, if n is a natural number other than 1 , then « — 1 is a natural number as well. Let S = {« : P{n) is false}. We want to prove that S is empty. Suppose that S is not empty. Then from theorem 3.5 it follows that S has a minimum m. Since P(l) is true, 1 (fc S. Hence m =fc 1 . Therefore m — \ is a natural number. But m is the minimum of S. Thus m — 1 S and so P(m — 1) is true. But then P(m) is true (because of hypothesis (ii) of the theorem). Hence m $ S which is a contra- diction. Natural numbers 23 3.9 Example For each n£N 1 +2 + 3 + . . .+« = \n(n + 1). Proof Let S„ = 1 +2 + 3 + ... + «= 2£ =1 k. Let P(n) be the state- ment S n = \n{n + 1). We have S 1 = l,and \. 1(1 + 1) = 1. Hence P(l) is true. We now wish to show that if P{n) is true, then P(n + 1) is true'. To do this, we assume that P(ri) is true and try and deduce that P{n + 1) is true. Thus we assume that S„ = \n(n + 1) and try and deduce that 5 n+1 = k(n + 1)(" + 2). But S n + l = 1 + 2 + 3 + . . . + n + (n + 1) = S n + (n + l) = \n(n + 1) + (« + 1) = &« + l)(n + 2) as required. The result now follows by induction. 3.10 Example The real number A n = —(at +a 2 + ■ • - + a n ) n is the arithmetic mean of the numbers a 1 ,a 2 , ... ,a„. If 01,22, ... ,a n are positive, their geometric mean G n is defined by G n = (a,a 2 ...fl n ) 1/n . The 'inequality of the arithmetic and geometric means' asserts that G n <A n . Proof This inequality has an entertaining proof using 'backwards induction' (see exercise 3.1 1(5)). LslP(n)be the statement 'For any positive numbers at, a 2 , . . . ,a n , G„ < A„'. We shall show that (i) P(2 n ) is true for each nGN. (ii) If P(n) is true, then P(n — 1) is true. The result then follows by backwards induction. (i) We begin by proving P(2). Now, < (V^i - y/a 2 ) 2 -at — 2yJ(a x a 2 ) + a 2 ( fl|a2 ) 1/2 <K«i+^)- This is simply P(2). We now show that P(2 n ) implies P(2" * '). Let m = 2". Then 2«+i _ 2m. We have to assume P(m) and try and deduce P(2m). Since P(m) is true, 24 Natural numbers (a 1 a 2 ...a m ) l " n <-(a 1 +a 2 1 •o and ( a m*\ a m+2 • ■ - a 2m) < ~~ ( a m + 1 + "m + 2 + • ■ • + «2m)- But we know that P(2) is true. Hence {( flia2 ...0 ,/m (^^m + 2---«2 m ) 1/m }" 2 ^ 1 (a, + . . . + a m a m+1 +...+a 2m ■ + m m Thus ( ai a 2 . ..a 2m ) V2m < — (a, + <z 2 4- . . . +a 2m ) 2ffi which is /'(2/m). We have shown that P(2 1 ) is true and that, if P(2 n ) is true, then P(2 n+1 ) is true. Hence P(2") is true for all n S |^J by straightforward induction. (ii) We now show that, if P(n) is true, then P(n — 1) is true. Suppose that P(n) is true. Then {(ha i ...a n . t G n - l y n < vn ^ai+a 2 + ... + a„_, + G„_i -»-] r MM / (H-lMn-l+Cn-1 ft G n _, </!„.,. ThusP(n — l)is true. This completes the proof. 3.11 Exercise (1) Prove by induction that n (i) I & 2 = l 2 + 2 2 + . . . + n 2 = fe = i (ii) T A 3 = l 3 + 2 3 + . . . + n 3 (2) Prove by induction that, if x =£ 1, £* fe = 1+x+x 2 + ...+*" fe=o in(n + l)(2n+I) i« 2 (« + l) 2 . l-x n+1 1-as Natural numbers 25 Deduce that, for any real numbers a and b, a"-b" m (a-bXa"' 1 + a"' 2 b + a"' 3 b 2 + . . .+ ab"' 2 + b"' 1 ). (3) A polynomial P(x) is an expression of the form P(x) = a n x n + a^iX"' 1 + . . . + aix+a . If a n =£ 0, the polynomial is said to be of degree n. Suppose that P(x) is a polynomial of degree n and that, for some if, P{£) = 0. Prove that P(x) = (x - %)Q(x) where Q(x) is a polynomial of degree n — 1 . [Hint: use the previous question.] If gj, 2- 2 , . . . , £„ are distinct real numbers and P{%\) = P(£ 2 ) = . . . = P(% n ) = 0, prove that P(x) = k(x-^)(x-$ 2 )...(x-U where A: is a constant. (4) Assuming that w! = 1.2.3...n and 0! = 1, we define m r\(n-r)\ Prove that (r=0,l,2,...,ii). «+l r (r=l,2,...,n). Using this result and the principle of induction, obtain the binomial theorem in the form {a + bf = a" + na"' 1 b + ^^ a"- 2 b 2 + . . . + b" = y. V a"- r b r . (5) Let P(n) be a statement about the natural number n and suppose that (i) P(2") is true for each n G M (ii) if P(n) is true, then P(n — 1) is true. Prove that P(n) is true for every n€N. [Hint: the set {2" : n S M } is unbounded above (exercise 3. 6(2)). J (6) Suppose that < X< Y and that x and y belong to the interval [X, Y]. Prove that, for each nGM, AT"" | jc -y K «^TK| x 1/n -/'" | < YX 1 '" \x-y\. [Hint: exercise 3.1 1(2).] Convergent sequences 27 = !{1 + (" £)+(-£)*+... + (-|f-*} CONVERGENT SEQUENCES 4.1 The bulldozers and the bee Two bulldozers are moving towards each other at a speed of one mile per hour on a collision course. When they started they were one mile apart and a bee was perched on the front of one of them. The bee began to fly back and for- ward between the bulldozers at a constant speed of two miles per hour, vainly seeking to avoid his fate. How far from his starting point will the unfortunate insect be crushed? fef" fe# fr^S This riddle is usually posed in the hope that the victim will embark on some complicated calculations involving the flight of the bee. But it is quite obvious that the bee will be crushed when the bulldozers collide. Since the bulldozers travel at the same speed, this will happen halfway between their starting points. The answer is therefore one half mile. Let us, however, take up the role of the riddler's victim and examine the flight of the bee. Letx„ denote the distance the bee is from his starting point when he makes his n th landing. Then r _ 2 x \ — 3 v - 2_2 x 2 — 3 9 and, in general. = §-!+f 7 -... + (-iy- 1 2(i)" 26 2 li -(-!)" 3 ll _ ( _, )1 =^a-(-r). How is the answer \ to be extracted from this formula? It seems clear that, as n gets larger and larger, (— £)" gets closer and closer to zero and so x n gets closer and closer to \. We say that 'x n tends to § as n tends to infinity' - or, in symbols, l x n -> \ as n -*■ °°\ This idea is of the greatest importance, but, before we can make proper use of it, it is necessary to give a precise formulation of what it means to say that x n -> I as n -* °°. 4.2 Sequences A sequence may be regarded as a list of numbers Xi,X2,x 3 ,x 4 , . . . More precisely, we can say that a sequence is determined by a rule which assigns to each natural number n a unique real number x n . We call x n the n th term of the sequence. The notation means the sequence whose «th term isx„. The set {x n : n € N} is called the range of the sequence. We say that a sequence is bounded above (or below) if its range is bounded above (or below). Thus (x„) is bounded above by //if and only if-xr„ <//(« = 1, 2, 3, . . .). 4.3 Examples (i)<(-ir> = -i,+ i,-i, + i,... oo<«- 1 >= i,|, M.. (iii)<l> = 1,1,1,1,... (iv)(2 n > = 2,4,8,16, (v) The sequence (x„) defined inductively by *i = 2 and 1 (n = 2,3,4, ...). The first few terms are 2, i. 12. I§> • ■ • X "~ 2Y n - l+ x n . r 4.4 Definition of convergence A sequence (x n ) is said to converge to the limit I if and only if the following criterion is satisfied. 28 Convergent sequences Given any e > 0, we can find an N such that, for any n>N, \x n -l\<e. We write x n ■* I as n ■* °° or lim x„ = /. It may be helpful to think o{x h x 2 ,x 3 , ... as successive approximations to the number /. The distance \x n —l\ between x n and / is then the error involved in approximating / by x„. The definition of convergence then simply asserts that we can make this error as small as we choose by taking n large enough. The diagram is supposed to represent a sequence <x„> with the property that x„ -*■ I as n -*• <*>. n>6 For the value of e indicated in the diagram, a suitable value for N in the definition isN= 6. For each value of« >6, \x n -l\<e. In particular, the value of \x 9 — l\ has been noted in the diagram. Notice that the definition of convergence begins 'Given any e > 0, we can find an N . . . ' Here the emphasis is on the very small values of e > 0. It is clear from the diagram above that, if we had chosen to look at a very much smaller value of e > 0, then we should have had to have picked out a very much larger value of N. In general, the smaller the value of e > 0, the bigger must be the corresponding value of iV. 4.5 Example 1 1 + — -* 1 as n -*■ °°. n Proof Let e > be given. We must find a value of ./V such that, for any n>N, Convergent sequences 29 But 1 »'" «'" <e. and so we simply need to find an N such that, for any n>N, i.e. -<6 n e But this solves the problem. We simply chose A' = 1/e. Then, for any n>N, i.e. i.e. n>N = - e -<e n l+ p. <e. We have shown that, given any e > 0, we can find an N (namely N=l/e) such that, for any n>N, <e. It is important to observe that N depends on e, i.e. for each e > we use a different N. Some values of e and the corresponding values of N are entered in the table below. e N 0-4 2-5 0-1 10 0000 001 1 000 000 (Note: Some authors insist that A' be a natural number. This makes the defi- nition of convergence a little more elegant but renders examples like that above marginally more complicated. If we wanted A' to be a natural number in the example above, we could not simply write A' = 1/e. Instead we should have to choose N to be some natural number larger than 1/e.) 30 Convergent sequences 4.6 Exercise (1) Prove that r-fc4i = n-"~ \n 2 + 1 , (2) Let r be any positive rational number. Prove that — -* as n •* °°. n (3) Let X be any real number. If x n -* I as n -> °°, prove that Xx n -> X/ as 4.7 Criteria for convergence In the example and exercises above it is fairly easy to decide what value of A' is appropriate to any given value of e > 0. But this is by no means always the case. It is therefore natural to look around for some shortcuts which will enable us to determine whether a sequence converges (and what its limit is) with- out our having to indulge in the painful process of appealing to the definition. In this and the next section we give some useful results of this sort. 4. 8 Proposition (combination theorem) Let x„ ■* / as n -* °° and y n -*m as n •* °° and let X and /j be any real numbers. Then (i) Xx n + ny n -*Xl + [xm as n -* °° (ii) x n y n -* Im as n -* °° x I (iii) — -*■ — as n ■* °° (provided that in i= 0). y n rn The proofs of (ii) and (iii) can be found in the appendix. These proofs are somewhat technical and we prefer not to hold up the discussion by presenting them at this stage. Readers who prefer to omit the proofs of (ii) and (iii) altogether will not suffer greatly as these results may be deduced from (i) once the theory of the exponential and logarithm functions has been developed. Proof of 4.8(ij. After exercise 4.6(3) we need only show that, if x n -* I aSH -+ °° and y n -*m asn-*°°, thenx n + y n ->l + m asn -»°°. Let e > be given. Then |e > 0. Since x n -* I as n ■* °°, it follows that we can find an A' 1 such that, for any n >N U \x n -l\<\e. (1) Similarly, we can find an A^ such that, for any n>N 2 , \y n -m\<\e. (2) Convergent sequences 31 Let Af be whichever of N t and Af 2 is the larger, i.e. A' = max {A^, A^}- Then, if n >N, both inequalities (1) and (2) are true simultaneously. Thus, for any n>N, \(x„+yn) -(/ + «)! = \(x n -l) + (y n -m)\ < \x n —l\ + \y n —m\ (triangle inequality) <\e + \e = e. Given any e > we have found a value of N (namely N = max {N\, N 2 }) such that, for any n > N, \(x n +y„) — (I + m)\< e. Hence x n +j„-»; + masn^». 4.9 Example Prove that / 2n 3 - 3ra .»?- ( 5 „3 + 4« 2 -2 Proof We write 2n 3 -2n 2-3/T 2 2-0 2 = — as n -* °°. 5n 3 + 4n 2 -2 5 + 4«-'-2« -3 5 + 0-0 5 The supporting argument is as follows. It is obvious from the definition of con- vergence that 2 -* 2 as n •* °°. From exercise 4.6(2), n' 2 -*■ as n •* °°. Hence, by proposition 4.8(i), 2-3w" 2 ^-2-3.0 = 2as«-"°°. Similarly, 5+4n- 1 -2M- 3 ^-5+4.0-2.0 = 5 as «->">=. The result then follows from proposition 4.8(iii). 4.10 Theorem (the sandwich theorem) Suppose that y n •* I as n •* °° and thatz n -»Z as«-*°°. Uy n <x n <z n (n = 1, 2, . . .), then x„ -* / as n ■* °°. (Here the sequence <^: n > is 'sandwiched' between the two sequences <>> n > and <z n > - just as the bee of §4.1 was 'sandwiched' between the bulldozers.) Proof For this proof it is necessary to note that the inequality \x — 1\ <eis true if and only if/ — e<x</+e. (See exercise 1.20(1).) l-e i ->♦«- / + , 7£ I.V-/I 32 Convergent sequences Let e > be given. We have to find an N such that, for any n>N, \x n -l\<e. Since y n -* I as n -> °°, we know that there exists an N t such that, for any n>N u \)' n -I\<e. (3) Similarly, since z„ -* / as n -* °°, there exists an N 2 such that, for any n >N 2 , \z n -l\<e. (4) Let A' be whichever of N t and A^ is the larger, i.e. TV = max {JVj.A'j}. Then, if n >N, both inequalities (3) and (4) are true simultaneously. Thus, for any n>N, l-e<y n <l + e and / — e<z n </ + e. But_y„ <x„ <z„ (n = 1, 2, . . .). Hence, for any n >N, !-e<y n <x n ^z n <l + e. Hence / — e<x n <l + e i.e. \x„—l\<e. Given any e > 0, we have found a value of A' (namely JV = max {N%, N 2 }) such that, for any n>N, \x n -l\<e. Hence x n -*■ I as n -* °°. 4.11 Corollary Suppose that v„ -» as n -> °° and that \x n -l\<y n («=l,2,...). Then x n -* I as n ■+ °°. Proof The inequality \x n — 1\ < y n is equivalent to / — y n < x n < / + y n . But, using proposition 4.8, / — y n -> / as n -» 0° and / + _y„ ■+ / as « -»«>. Hence x n -> / as n ■* °°, by theorem 4.10. 4.72 Example Suppose that |.v| < 1. Prove that x" -»• as « -» °°. TVoo/ Write . .. 1 1 xr = l < — . Thus (!+/?)" 1 + nh + * 2 n(n - \)h 2 + . . .+h" nh Convergent sequences 33 \x n -Q\ = \x\ n <— («= 1,2,.. .). But 1/rt -> as « -*■ °° (by exercise 4.6(2)). Thus, by corollary 4.1 1 , x" •* as n -*■ °°. 4.13 Example In the problem of the bulldozers and the bee, we obtained the sequence By example 4. 1 2, (— J)" -> as « -*■ °° and hence x„ -» \ as w •* °°. 4.74 Example Let x > 0. Prove that x m ■* 1 as n -> °°. />oo/ In the inequality of the arithmetic and geometric means, take 1 = a , = fl:2 = . . . = a n _, and a„ = .*. Then G„ = x 1/n and>4 n = (n — 1 + *)/« (x-l)/n + 1. y — 1 Hence x 1,n < +1 i.e. : i/"_! < _A ( _iy (Alternatively, take X =y = 1 in exercise 3.1 1(6).) Assume to begin with that x > 1 . Then 0<x 1/n -l <-(x-l) and hence *"" -* 1 as n ■+ °° by the sandwich theorem. If < x < 1 , then x =y~ l where y > 1. Buty 1 '"-*- 1 as w -► o °and hence 1 1 by proposition 4.8(iii). 4.15 Monotone sequences A sequence Cc„> is increasing if x n+l >x„ (n = l,2, ...). 34 Convergent sequences Similarly, <x n > is decreasing if X n+l <X„ (rt=l,2 ). (.v., > increases (v„ ) decreases A sequence Oc„> is strictly increasing if x n + 1 > x n (n = 1 , 2, . . .). It is strictly decreasing if x n + l < x n (n = 1 , 2, . . .). A sequence which is either increasing or else decreasing is called monotone. 4.16 Example (i) The sequence (2"> is strictly increasing. (ii) The sequences (n~ l ) and (— n) are both strictly decreasing. (iii) The sequence (1) is both increasing and decreasing. (iv) The sequence {(— l) n > is not monotone. It is neither increasing nor decreasing. 4.17 Theorem (i) If <.x„) is increasing and bounded above, then it converges to its smallest upper bound. (ii) If <*„> is decreasing and bounded below, then it converges to its largest lower bound. Proof If (x„) is decreasing and bounded below, then <— x n ) is increasing and bounded above. Hence it is only necessary to prove (i). Suppose that (,x n ) increases and is bounded above with smallest upper bound B. We must show that x n -*■ B as n -*■ °°. Let e > be given. Since B — e is not an upper bound, there exists at least one term x N of the sequence such that x N >B-e. But (x„) increases. Hence, for any n >N,x„ ~>x^. Thus, for any n>N, x„>x N >B — e. Convergent sequences *S Also B is an upper bound for the sequence. Hence x„ < B (n = 1 , 2, . . .). Thus, for any n>/V, B-e<x n <B. B-e<x n <B + e i.e. \x n -B\<e. Given any e>0, we have found an Af such that, for any n >N, \x n — B\<e. Hence x n -* B as n -*■ <*>. 4.18 Examples (i) The sequence <(« - l)/«) increases and we know from exercise 3.6(1) that it is bounded above with smallest upper bound 1 . By theorem 4.17 it follows that «-l 1 as n -*■ <*>. (ii) If < x < 1 , the sequence Ct"> decreases and is bounded below with largest lower bound (exercise 3.6(2)). Thus as n 4.19 Example The examples above are nothing new. But consider the sequence ((] +«-')">. We first show that this sequence increases. In the inequality for the geometric and arithmetic means, take a^ — a 2 = ■ ■ ■ — a n -\ = 1 + (« — 0" and a n = 1 . Then 36 Convergent sequences Hence x | i y"- i ^ < (^-ixi + (/2 -in + 1 = [j + i 1 +■ n-1 <ll+4l §1-2,3,...) and so the sequence increases. To be able to deduce the convergence of the sequence from theorem 4.17, we also need to prove that the sequence is bounded above . By the binomial theorem, 'l\ n(n-\)(\ U ,+ b ■"•l? SI- M; : ■ -2i-h+ H H)h- " \ nl <I + 1 U + ±+...+l 2! 3! r! 1 - «-l\ 1 n I n\ 1 1 1 < 1 + 1 +-+-^+ .. . + -^r, (because 2" _1 <«!) = 1+: tz? = i +2(1- an < 3. Hence the sequence is bounded above by 3. From theorem 4.17 it follows that ((I + aT 1 )"} converges to a limit B and B < 3. (In fact, B = e = 2-718 . . .) 4.20 Exercise (1) Prove that n 3 + Sn 2 + 2 . 2n3 + 9 -in*--. (2) Decide for what values of x the limit [x+x n \ lim 1 + x n exists. Draw a graph which plots the value of the limit against the value of*. (3) Use the sandwich theorem to prove that y/(n + 1) — y/n -> as n -*■ °°. Convergent sequences 37 " (4) Let x be a positive number, and let N be the smallest natural number satisfying N>x. Prove that ^< x"- n\ (N-l)l \N n-N+l (n > N). Deduce that x n /n ! -*■ as n -* °°. ' (5) Let a be any positive rational number and let \x\ < 1 . Show that there exists a natural number N such that (1 + 1/A0 a+1 |»l < 1. Deduce that 1 x"\<\N a * 1 x a + l„iV| (n>N). Hence show that n a x n ■+ as n -*• °°. '(6) Prove that the sequence (n Vn ) decreases for B>3. [Hint: example 4.19.] Hence prove that the sequence converges. (The convergence of <n 1/n > can also be established by a method like that of example 4.12. If n = (1 + /*„)", then it follows from the binom- ial theorem that n -^Hl<n. Therefore h n •* as « -* °°.) 4.21 Some simple properties of convergent sequences 4.22 Theorem A sequence can have at most one limit. Proof Suppose that x n -* I as n -» °° and x n -> m as « -> °°. Let e > be given. Then |f— »| = \l-x n +x n -m\<\l-x n \ + \x n -m\<e + e = 2e (triangle inequality) provided that n is sufficiently large. But, from example 1 .7 it follows that, if < ||/ - m\< e for every e > 0, then |/ - m\ = 0, i.e. l = m. 4.23 Theorem Suppose that x„ ■+ / as n ■* °°. (i) If x„ > a (n = 1 , 2, . . .), then / > a. (ii) If x n <b(n = \,2,.. ), then / < b. Proof Kx n <b(n= 1,2,. . .), then -x„ >-b (n = 1, 2, we need only prove (i). Let e > 0. Then there exists an N such that, for any n>N, \x„-l\<e i.e. l-e<x n <l+e. .). Hence 38 Convergent sequences But x n > a (n = 1 , 2, . . .) and so, for any n > N, a < x n < I + e. Hence, given any e > 0, a < / + e. From example 1 .7 it follows that a <. I. 4.24 Example Suppose that x n ■* I as n -* °°. If x n > a (« = 1 , 2, . . .), it is tempting to conclude that l> a. But this may not be true. For example, l/n -*■ as n -*■ °° and l/n > (n = 1 , 2, . . .). But we certainly cannot conclude that 0>0. 4.25 Tiieorem Any convergent sequence is bounded. Proof Let x n -*■ I as n -*■ <*>. We have to find a ^ such that |jc„| < K (n = 1, 2, . . .). (See proposition 2.3.) It is true that, for any e > 0, there exists an N such that, for any n>N, \x„ — 1\ < e. In particular, this is true when e = 1, i.e. there exists an N t such that, for any n> JVj, Ix„-/I<l. From theorem 1 .18 it follows that, for any « > A 7 !, |jc b I-|/|<I*„-/|<i i.e. M<|l| + ] (n>iV,). The result now follows if we take K = max{|.r,|, \x 2 \, ..., |jf»|, |/| + 1}. 4.26 Divergent sequences A divergent sequence is a sequence which does not converge. Any un- bounded sequence is therefore divergent (theorem 4.25). However bounded divergent sequences exist as well. 4.27 Example The sequence <(— l) n > diverges. Proof The sequence is obviously bounded. Suppose that (— 1)" ■* / as n -*■ °°. Then, given any e > 0, we can find an N such that, for any n> N, l(~ 1 )" — /| < e. But there are both even and odd values of n greater than N and so 1 1 — /|< e and |— 1 — /| < e. These inequalities must be true for any e > and so we obtain the contradiction / = 1 and / = — 1 . We say that a sequence (x n ) diverges to + °° and write x n -*■ + °° as n-* °° if, for any H > 0, we can find an IV such that, for any n>N, Convergent sequences 39 x n >H. Similarly, a sequence <x n ) diverges to—°° and x n •* — °° as n -*■ °° if, for any H > 0, we can find an iV such that, for any n > N, x n <-H II - N - :i< r. - -- x„— *— «asn— * A divergent sequence <A" n > which does not satisfy x n -> + °° as n ■* °° or *,,->• — ooas«->°°is said to oscillate. An example is the sequence <(— 1)">. 4.28 Example Let a be a positive rational number. Prove that if -* + °° as n -*°°. n>N, Proof Let H > be given. We have to find an A' such that, for any n a >H i.e. n>H ila . We simply choose A' = H Ua . Then, for any n > N, n >H va and hence n a > H. A common error is to suppose that proposition 4.8 (about combining con- vergent sequences) applies to sequences which diverge to + °° or — °°. For example, it is tempting to suppose that, if x n -* °° as n •* °° and y n •* °° as n •* °°, then x n —y n -»°° — °° = 0asn->°°. But °° is not a real number and cannot be treated as such. Indeed, one should immediately be warned that something is wrong by the appearance of the totally meaningless expression °° — °°. To press the point home, we consider the example x n = n 2 (n = 1, 2, . . .) andy„ = n (n = 1,2,...). Then x n -*• °° as n •* °° and y n -*■ °° as n -> °°, but x n - .y n "* °° as « ->°°. Or, again, takex„ = (n + 1) (n = 1, 2, . . .) andy n = n (n = 1, 2, . . .). This time jr n — y n -*■ 1 as « -* °°. 40 Convergent sequences 4.29 Exercise (1) Suppose that x n -*■ I as n -> °°. Prove that (i) \x„ — 1\ -> as n -*■ °° (ii) \x n \ ■* |/| as« -*«>. (2) Suppose that x n -*■ / as « -+ °°. If / > 0, prove that there exists an N such that, for any n >N, x n >\l. (3) Prove that (i) 2" -*■ + °° as n -* °° (ii) — V" ""* ~~ °° as " "* °° (iii)<(-l)"n> oscillates. (4) Let jc„ >0 (« = 1 , 2, . . .). Prove that x„ -» as « ■* °° if and only if l/x n -*-°°as/! ^°°. (5) Suppose that {x n ) increases and is unbounded above. Prove that x n -*■ + °° as n ■* °°. If <x n > decreases and is unbounded below, prove that x n -> — °° as « * °°. * (6) Let 5 be a non-empty set of real numbers and suppose that d{%, S) = (see exercise 2.13(4)). Show that, for each n £ N, we can find an x n GS such that |£ — x n \ < l/«. Deduce that x n -> i- as « -> °°. If S is bounded above, show that a sequence of points of S can be found which converges to its supremum. If S is unbounded above, show that a sequence of points of S can be found which diverges to + °°. SUBSEQUENCES 5.1 Subsequences Suppose that < x n ) is a sequence and that < n r > is a strictly increasing sequence of natural numbers. Then the sequence is called a subsequence of < x n ). The first few terms of some subsequences of < x n > are listed below. \X r+i ) = ^2,^3,^4, . . . '*3r ' = x 2> x 4> x 6> ■ ■ ■ '*3r+S' = x S> x ll> x 14i x n> ■ ■ ■ (X^r) = X2, Xn, -Xg, Xi6, . . . Roughly speaking, if we think of a sequence as a list of its terms, then we obtain a subsequence by crossing out some of the terms. For example, < x 2n ) may be obtained from <x n >by crossing out terms as below. Al> x 2> A3. • JC 4>/S' *6> ■ • ■ The insistence that < n r > is strictly increasing is of some importance. It means, for example, that the sequence whose first few terms are -*3> X \, X S, X \> X <)> ■ • • is not a subsequence of < x n ). The fact that (n r ) is strictly increasing has the consequence that n r >r (r = 1,2, .. .). This is easily proved by induction. 5.2 Theorem Suppose that x n -»/ as n-*°° and that <x„ r > is a subsequence of (x n ). Then x n -* I as r •* °°. Proof Let e > be given. Since jc„ ■* I as n -* <*>, there exists an iVsuch that, for any n >N, 41 42 Subsequences I *,-/!<«. Lei R=N. Then, for any r>R,n r >r> R = N and so w r > N. Thus We have shown that, given any e > 0, we can find an R (namely R = N) such that, for any r>R, \x„ F -l\<e. Hence x„ ■* I as /■ ■* °°. 5. J Example We have seen that the bounded sequence ((— 1)"> diverges. A simpler proof can be based on theorem 5.2. Obviously the subsequence of odd terms tends to — 1 and the subsequence of even terms to + 1. But, if <(— 1)" > converged, all its subsequences would tend to the same limit. 5.4 Example In example 4.4 we showed that, if x > 0, then x vn •* 1 as n -*■ °°. We prove this again by a different method. As in example 4.4 we need only consider the case x>l. If x > 1 , then x lln > 1 (n = 1 , 2, . . .) and hence the sequence < x lln ) is bounded below by 1 . Moreover I/n; l/(n+l) _ (n+l-n)/n(n+l) _ "/* = X = x 1/n(n+1) >l (« = 1,2,...) and therefore (x' ,n ) decreases. From theorem 4.17 it follows that <*""> con- verges to a limit / and / > 1 . What is the value of /? By theorem 5.2, we know that all subsequences of (x 11 ") must also tend to /. Hence ,1/2" -*• / as n -*■ °°. Using proposition 4.8, it follows that But a sequence can only have one limit. Thus / = I 2 and so / = or / = 1 . But / > 1 and hence / = 1 . We have shown that Aln -*■ 1 as n -*■< Note: A warning is appropriate here. Do not use this method of working out the value of a limit until you have shown that the sequence converges. For example, if you thoughtlessly tried to evaluate the limit of the sequence <(-- 1)"> by looking at its subsequences, you would obtain / = + 1 = — 1. ' Subsequences 43 5.5 Example Let a >0 and let jc, >0. Define the rest of the sequence <x n ) inductively by *„ +1 = Kx n + OX?) in = 1,2,...). Obviously x n >0(n = 1,2, . . .). But further x 2 n -2x Ml x n +a = 0. We know in advance that this quadratic equation in x n has a real solution. Thus 'b 2 - 4ac > 0' , i.e. x 2 +l >a. Since jc n+1 > 0, it follows that x n+l >s/a (n = 1,2,...). We wish to prove that < x„ > decreases and so we consider * n -*n+i = x„-i(x„+ ax- 1 ) > (n = 2,3,...). Hence the sequence <jc„> decreases and is bounded below by Va (provided we omit its first term). From theorem 4.1 7 it follows that x n -* I as n -* °° where / > \Ja. What is the value of/? By theorem 5.2 we also have x n+1 -* / as n -* °°. But Now / ^ (because / > \Jd). Hence, by proposition 4.8, x n +i = i(* n + ^n 1 )- > K' + «'" , )asn^ 00 - Since a sequence can have at most one limit, / = kQ + al" 1 ) i.e. I 2 = a. Thus / = yJaoxl = — \Ja. But we know that / > \/a. Hence / = \/a. We have therefore shown that x n -* \Ja as n -* °°. 44 Subsequences 5.6 Example We attack the previous example by a different method. Again let a > but this time let Xi be unspecified for the moment. Define the rest of the sequence inductively by Now x n +i = ttxn+ax- 1 ) (n = 1,2,...). x n+l ~\/a = \(x n + ax~ l )-s/a 2x, (x 2 n -2x n y/a + a) ' 2x n \ 2jc n _, 1 1 2x n (2* n _,) 2 {x M - Va) 4 1 1 1 1 2x n (2x n „ 1 ) 2 ---(2x i y P5<*i-v«r. If we assume that Xj > \/a, then it follows, as in example 5.5, that x n > y/a (n= 1,2, . . .). Hence l* n +i-\/«l < 2%Ja 2y/a l + 2+2 2 +...+2" _1 c^-v«r (2"-l)/(2-l) (*i-vfc)* n = 2Va ■ i~ Va 2Va We know that, if 1 7 | < 1 , theny" -> as « ■+ °°. By theorem 5.2 it follows thaty 2 -*0asn-*°°. Hence our argument shows that x n •* \Jaa.sn^>°° provided that 2\/« <1. We have already assumed that x x > \Ja and so we have shown that x n -* V<z as « -* °° provided that \]a <x y < 3 \Ja. Of course we already know from example 5.5 that this was true under the weaker hypothesis that x x > 0. However, the Subsequences 45 argument given here is interesting because it allows us to estimate how 'good' an approximation x n is to \/a. For example, suppose we want to estimate the value of y/2. It is obvious that 1 < sj2 < 2 (because l 2 = 1< 2 and 2 2 = 4 > 2). For a better estimate take a = 2 and*! = 2 in the argument above. Then |x 4 -V2|«2V2 2V2 (2-V2 2V2 = 2\/2 V2(V2 - 1) 2V2 (V2-1) 8 . Using the crude estimate \J2< 2 we obtain |x 4 — \/2 |<2 _6 = p< 0-016. Hence x 4 = gg = 1 414 . . . differs from y/2 by at most 0016. (A better estimate for \J2 may be obtained by evaluating x s or by starting with Xj = \ and using the estimate \J2 < §.) 5. 7 Exercise (1) Given that < n u " > converges (see exercise 4.20(6)), show that /i l/n -> 1 asH-* 00 by considering the subsequence ((2m) 1 ' 2 "}. (2) A sequence < x n > is defined by x x = h and x n*l = x n ' " where < k < \ and h lies between the roots a and b of the equation x 2_ x+k=Q Prove that a <x m . 1 <X n < b {n = 1, 2, . . .). Show that (x n ) con- verges and determine its limit. * (3) If k > and x x > and ( x n > is defined inductively by ■*n+1 — l+x n ' show that one of the sequences <,x 2n > and (x 2n -i > is increasing and the other decreasing. Prove that both sequences converge to the limit / which is the posi- tive root of the equation x 2 + x = A:. What conclusion may be drawn concerning <*„>? ' (4) Two sequences < x n > and (y„ > are defined inductively by x l = \ and y-i = 1 and X„ = VCXn-lJ'n-l) (» = 2,3,...) 46 Subsequences 2 W„ J. y n -i) (n o 2,3,...)- Prove that x„_, <x„ <^ n <>>„_, (« = 2, 3, . . .) and deduce that both sequences converge to the same limit /, where ? < / < 1 . (Actually / = ff/4.) 1 (5) Let y be any real number. Prove that the sequence <(l+yn- l ) n ) is increasing for those values of n which satisfy n > 1 —y. [Hint: example 4.19.] Show also that the sequence is bounded above and hence converges. [Hint: From exercise 4.20(4), (2y)"/n\ ->0 as n -> °° and therefore we can find an N such that, for any n>N,y"/nl< (£)".] ' (6) Show that the product of the limits of the two sequences <(1 + xn' 1 )") and <(1 -xn'')") is equal to 1. '" 1 + « V~7i i-$ n 2 \l/n-i 5.8 5.9 Bolzano-Weierstrass theorem Theorem Every sequence has a monotone subsequence. Proof Let (x n > be any sequence of real numbers. We must construct a subsequence {x„ r ) which is either increasing or decreasing. We distinguish two cases. (i) Every set {x n : n > N} has a maximum. In this case we can find a sequence (n r ) of natural numbers such that x n = maxjc n n>l *„,, = maxx n n>n, x„ 3 = max x n and so on. Obviously n t <n 2 <n 3 < . . . and so, at each stage, we are taking the maximum of a smaller set than at the previous stage. Hence < x n > is a decreasing subsequence of < x n > . (ii) Suppose that it is not true that all of the sets {x n : n > N} have a maxi- mum. Then, for some jV, , the set {x n :n>N t } has no maximum. It follows that, given any x m with m>N lt w can find an x n following x m such that x n >x m . (Otherwise the biggest of x N +1 , . . . ,x m would be a maximum for {*„:«>AM). 47 Subsequences We define x ni = x Ni+1 and then let x„ 2 be the first term following *„, for which x n >x n '. Now let x nj be the first term following x„ 2 for which x n >x n . And so on. We obtain an increasing subsequence <*„ r > of (x n ). 5.10 Theorem {Bolzano-Weierstrass theorem). Every bounded sequence of real numbers has a convergent subsequence. Proof Let < x n > be a bounded sequence. By theorem 5.9, (x n > has a monotone subsequence <*„,.>. Since (x n ) is bounded, so is <x„ r >. Hence, by theorem 4.17, <x„ r > converges. 5.11 Examples (i) The sequence <(— 1)" > is bounded. Two convergent subsequences are the sequence — 1 , — 1 , — 1 , ... of odd terms and the sequence 1 , 1 , 1 , 1 , . . . of even terms. (ii) A more sophisticated example is the sequence ( r n > whose first few terms are . 2' 3> 1' 4> 4i $> S< ?i 5> 5> Sj • ■ • Every term of the sequence is a rational number between and 1 . Hence the sequence is bounded. It has many convergent subsequences. For example 5, 4> 6> S. • • • ill! 2> 3. 4> Sj ■ • ■ uu... 5.12* Lim sup and lim inf We shall not have a great deal of use for the subject matter of this section. However, it is material which has to be learned eventually and it is natural to deal with it here. Suppose that <jc„> is a bounded sequence and let L denote the set of all real numbers which are the limit of some subsequence of < x n >. We know from the Bolzano-Weierstrass theorem that L is not empty (i.e. L # 0). Of course, if (x„) converges, L will just consist of a single point (theorem 5.2). 5.13^ Proposition Let < x n > be a bounded sequence and let L be the set of all real numbers which are the limit of some subsequence of < x n >. Then L has a maximum and a minimum. That L is a bounded set follows from theorem 4.23. Hence L certainly has a supremum and infimum. Proposition 5.13 asserts that these actually belong to the set L. The proof is relegated to the appendix. Readers who prefer to omit 48 Subsequences the proof altogether will encounter it again at a later stage since it is a special case of a general theorem concerning 'closed sets'. If (x n > is a bounded sequence, let us denote the maximum of L by 7 and the minimum by /. Then / is the largest number to which a subsequence of <x„> converges and / is the smallest number to which a subsequence of < x n > con- verges. We call / the limit superior of ( x n > and write / = lim sup x n . n—*oo Similarly, / is the limit inferior of <x„ > and we write / = lim inf x„. (Occasionally you will see such expressions as lim sup x n = + <*>. This simply means that < x n > is unbounded above. Similarly lim inf x n = - °° means that (x n ) is unbounded below.) 5.14 * Examples (i) The sequence < 1/n > converges with limit 0. By theorem 5.2, all its sub- sequences converge to 0. Hence L contains the single point and lim sup — = lim inf — = lim — = 0. (ii) The sequence <(— 1)"> is bounded. The set L consists of two points + 1 and-1 (i.e.Z, = { + 1, - 1}). We have lim sup (-1)" e 1; lim inf (-1)" = -1. M,U, (hi) Let < r n > be the sequence whose first few terms are \, J, §, „, , I, \, . . . Obviously <r n < 1 (n = 1, 2, . . .). From theorem 4.23 it follows that L is a subset of [0, 1 ] (i.e. L C [0, 1 J). But the subsequence \, 1,1,1,... con- verges to 0. Thus / = 0. Also the subsequence £, \, |, f, . . . converges to 1 . Thus / = 1 . We have shown that lim sup r n = 1 ; lim inf r n = 5.15* Exercise *(!) Calculate (i)lim sup {(-!)"(!+ «-»)} (ii) lim inf {(-1)"(1 + «"')} Subsequences and show that these are not the same as the numbers 49 sup (-1 )"(! + « _1 ) and inf (- 1)"(1 »-»). * (2) Let < r n > be the sequence whose first few terms ate §, hhk, $»t*h 5* I* |,4» .. . Show that every point in the interval [0, 1] is the limit of a subsequence of <r„>. [Hint: see exercise 3.6(4).] t (3) Suppose that (x n ) is a bounded sequence and that, for any N, we can find an n > N such that x n > b. Show that < x n > has a subsequence which converges to a limit / > b. * (4) Let < x„ ) be a bounded sequence with limit superior / and limit inferior /. Show that, given any e > 0, we can find an N such that, for any n>N,x n <l + e. [Hint : use question 3 with b = / + e] . What is the corresponding result for /? t (5) Deduce from question 4 that a bounded sequence < x n ) converges with limit / if and only if lim sup x n = lim inf x n = I. Hence show that a bounded sequence < x n > converges if and only if all its convergent subsequences have the same limit. * (6) Let < x„ > be a bounded sequence and let ,M n = sup x k . Show that < M n ) decreases and is bounded below. Deduce that (M n ) converges and denote its limit by M. If / is the limit of some subse- _ quence of < x n >, show that M n > I (n = 1 , 2, . . .). Deduce that M > I. Obtain also the reverse inequality M < I [Hint: use question 4] and hence show that lim sup*,, = lim {sup x k }. (This explains the choice of notation for the limit superior. We also have, of course, lim inf x n = lim { inf x k } .) 5.16 Cauchy sequences Suppose that we want to prove that a sequence < x n > converges but we have no idea in advance what its limit / might be. Then there is no point in appealing directly to the definition because we shall certainly not be able to prove that \x„ — I \ can be made as small as we choose by taking n sufficiently large if we do not know the value of /. 50 Subsequences Theorem 4.17 gives us one way of resolving this difficulty. It tells us that an increasing sequence which is bounded above converges and that a decreasing sequence which is bounded below converges. Thus we are able to deduce the convergence of the sequence without necessarily knowing the value of its limit. But, of course, theorem 4.17 deals only with monotone sequences. In order to deal with sequences which may not be monotone, we introduce the idea of a Cauchy sequence. We say that < x n ) is a Cauchy sequence if, given any e > 0, we can find an N such that, for any m >iVand any n>N, \x m —X n \<e. Very roughly, the terms of a Cauchy sequence get 'closer and closer' together. 5.17 Proposition Any convergent sequence is a Cauchy sequence. 5.18 Proposition Any Cauchy sequence is bounded. The proofs are easy. The next theorem is what makes Cauchy sequences important. 5.79 Theorem Every Cauchy sequence converges. Proof Let (x n ) be a Cauchy sequence. By proposition 5.18, (x n > is bounded. Hence, by the Bolzano-Weierstrass theorem, it has a convergent sub- sequence (x nr >. Suppose that x„ r -*■ / as r -*■ °°. We shall show that x n -* 1 as Let e > 0. Then \e > 0. Hence there exists an R such that, for any r >R, „ r -/K£e. 0) Since < x n > is a Cauchy sequence, there also exists an N such that, for any m>N and any n>N, \Xm-X n \<y. (2) Now suppose that n > N and choose r so large that n r >Nandr>R. Then (1) is satisfied and also (2) is satisfied with m = n r . Thus, for any n > N, \x n -l\ = l**-*„ r + ;c nr -/| < I x n — x„ r | + \x„ r — / 1 (triangle inequality) < ^e + ie= e. Given any e > 0, we have found an A'' such that, for any n > N, \x„ — 1 1 < e. Thus x n -> / as n -> °°. Subsequences 5.20 Example A sequence <x n ) is defined by x, = a, x 2 = b and x n + 2 = |(Xn+l + *n) (" = 1,2,3,...). Prove that <x n > converges. Proof We have X n *2~ x n+\ = iOn+l +X n) ~ x n-H = h( x n ~ x n+l)- 51 Thus \X„+2~ x n*l I a «l*« x n+\\ 2 |A " "-' — ^,2 ' X " X "-l 1 1 — \x 2 -x l \ = 2 n Hence, if n >m, \Xm—X»\ = \x n -x n ^+x n _,-x„_ 2 -.. .+x m+i -x ri *tl ••mi < I X n -X„_ t I + I X n _, -X n -2 i + • • ■ + I x m+l ~ x n i=i+=i+.-- + 5^: 2 m - 1 1 - 1 Let e > be given. Choose N so large that 1 '4 + ? + - + 5^' 1 rt",-,Kil»-,l. ,N-2 Lft —«l<«. Then, for any n > N and any m > N. Thus < x n > is a Cauchy sequence. Therefore, by theorem 5.19, it converges. (An alternative method would be to show that one of the two sequences <*2n-i> and <*2*> is increasing and the other decreasing and then to show that they have the same limit.) 5.21 Exercise (1 ) Suppose that < a < 1 and that < x n > is a sequence which satisfies I x n+i ~ x n I < a " (" ~ 1 ■ 2 - • • •)• Rmw that < x " ) is a Cauch y sequence and hence converges. 52 Subsequences Give an example of a sequence < y„ > such that y n ■* + oo as n _► oe but lj' n+1 ->' n |-*0asM-*oo. [Hint: see exercise 4.20]. (2) A sequence <x„> satisfies 0<a <x, <x 2 < b and * n+2 = {*n + i*„}" 2 (« = 1,2,...). Prove that a <x n < b (it = 1 , 2, . . .). Hence or otherwise show that 6 I I ***! -*»!<■ -*n -**3 (« = 2,3,...). Deduce that <*„> is a Cauchy sequence and hence converges. (3) For the sequence of example 5.20, prove that x n+i + hx n = x n + \x n _ i = ... = x 2 + \x { . Deduce that | x n+i -l\ = l\x n -l\, where / = |(jc 2 + £*i> What con- clusion may be drawn about the convergence of the sequence <*„>? Tackle exercise 5.21(2) by a similar method. (4) Let [a, b] be a compact interval (see §2.9). Prove that every sequence of points of [a, b] contains a subsequence which converges to a Doint of [a, b]. » (5) Let /be an interval which has the property that every sequence of points of /contains a subsequence which converges to a point of/. Prove that /is compact. [Hint: First prove that /is bounded by assum- ing otherwise and appealing to exercise 4.29(6). Then prove that sup / and inf /are both elements of /with another appeal to exercise 4.29(6).] T (6) Given a set S of real numbers, let 5j = {x:xGSandx=t$. We say that | is a cluster point (or point of accumulation or 'limit point') of £ if £ is at zero distance from S±. (Note that % need not be an element of S). A form of the Bolzano- Weierstrass theorem asserts that every bounded set with an infinite number of elements has at least one cluster point. Prove this. SERIES 6.1 Definitions Given a sequence <c n > of real numbers the sequence <s N ) defined by hf ~ L a n ~ fl l + fl 2 + • ■ • + "N is called the sequence of partial sums of the series Z a n . n = l If s N ■* s as N -» °°, the series is said to converge to the sum s . We write S = £ a n . n = l It is important to remember that this formula can only make sense when the series converges. 6.2 Example Consider the series n = o If x J= 1 , the partial sums satisfy Sfl = £ x n = 1 + x + x 2 + . . . + x n = N 1 -x .\'+ 1 1 -x If |x|< l,.r N+1 -*0 as A'-* °° and hence 53 54 Series Series 55 s N - 1 as N -*■ °°. 1 -* It follows that the series 2~ =0 x" converges if |jc| < 1 and we may write 1 !>" = l-x (W<1). If \x I > 1 , the series diverges. 6.3 Example The partial sums of the series CO J I — — n = i n(n+ 1) are given by -l|i- n = 1 n(n + 1) n = i \n n 4- 1 4WBWH+ l i N N+ 1 = 1 -• 1 --> 1 asAf-* 00 . N+ 1 Hence the series converges and we may write 1 n = \n(n + 1) = 1. 6.4 Series of positive terms Series whose terms are all positive (or non-negative) are particularly easy to deal with. This is because the sequence of partial sums of such a series is increasing. Thus, if we wish to show that a series of positive terms converges, we only need to show that its sequence of partial sums is bounded above. If the sequence of partial sums is unbounded above, then the series diverges to + °° (see exercise 4.29(5)). 6.5 Theorem The series n = i n diverges to + °°. Proof Since the partial sums of this series increase, we only need to show that they are unbounded above. But w s 2N = I+^ + t+.-. + ^v i 2 N 4 4 iN-l Thus the partial sums are unbounded above and the theorem follows. 6.6 Theorem Let a be a rational number such that a > 1 . Then the series GO 1 converges. Proof Since the partial sums are increasing, we need only show that they are bounded above. For TV > 1, 1 . 1 1 + i T(JV-l)a + . . .+ 1 (2 N -lfj < 1 ;<* 2° \4 a 4" 4" 4 a ! + 1 2<fi-i)u + ...+ 1 2(N-l)Of 9JV-I 2 4 = 1 + 2 a 1 + 1 2 «-l ■ ^or- al - 1\JV + . . .+ N-l l-(l/2 a - 1 )^l-(l/2 a - 1 ) 56 Series provided that l/2 a_1 < 1 . But this follows from the assumption that a > 1 . Hence <s^> is bounded above and the theorem follows. 6.7 Elementary properties of series 6.8 Theorem Suppose that the series 2„ = i a„ and S" =1 b n converge to a and respectively. Then, if X and p. are any real numbers, the series 2„ = i (ka n + pb n ) converges to A a + ju0. Proof We have IMA) = X la n +n £>„ n = l ,. = 1 n = l -+\a + p@ asiV-x* 5 by proposition 4.8. 6.9 Theorem Suppose that the series 2„ = i a n converges. Then a n -> as « ■* oo. Proof Let the sum of the series be x. Then N s n = Z a„-»-sas?j->°°. n = l Also x.v _ l -*■ s as w -* <*>. But then flw = (fli +a 2 + • • • + %) - (fli + a 2 + • • • + %-j) = s N — s N - l -*s— s = OasTV-*- 00 . 6.70 Examples The series X(-D" n = l diverges. This can be deduced from theorem 6.9 by observing that its terms do not tend to zero. Note that the converse of theorem 6.9 is false. Just because the terms of a series tend to zero it does not follow that the series converges. Theorem 6.5 provides an example. The terms of the series OS 1 I 1 n=i n Series 57 tend to zero, but the series diverges. One might say that the terms of the series do not tend to zero 'fast enough' to make the series converge. 6.11 Proposition Suppose that the series S" =1 a„ converges. Then, for each natural number TV, the series £„ = /v a n converges and CO Y, fl n -*0as7v'-»-°°. n=N This result is often referred to by saying that the 'tail' of a convergent series tends to zero. The proof is easy. 6.12 Series and Cauchy sequences In the series we have studied so far, we have either been blessed with a nice formula for the partial sums (examples 6.2 and 6.3) or else the partial sums increased and so we only had to consider whether or not the partial sums were bounded above (theorems 6.5 and 6.6). What should we do in the absence of such favourable conditions? We ask the question: is the sequence of partial sums a Cauchy sequence*! This question has some fruitful answers as we shall see be- low. 6.13 Theorem Suppose that <a„> is a decreasing sequence of positive num- bers such that a„ -> as n -*■ °°. Then the series oo converges. Proof We show that the sequence <s„> of partial sums of the series is a Cauchy sequence. From theorem 5.19, it then follows that the series converges. The proof depends on the fact that, for each n>m,we have the inequality 0<a m + 1 — A m + 2 +a m + 3 — . . .<2„<fl m + 1 . This follows easily from the fact thata*, — aj, + 1 is always non-negative because <«(,> decreases. Let e > be given. Since a n -*■ as n -* °° we can find an N such that, for any n>N,a n <e. But for any n>m>N, \$n~ *ml = lfai-«2 +a 3 -. . .a n )-(a,—a 2 + . . .a m )l = la m + i — a m+2 +a m+3 — . . .a„\ < fl m + i<e (because m >N). Thus <s„> is a Cauchy sequence and the theorem follows. 58 Series 6.14 Example We saw in theorem 6.5 that the series £™ =1 l/« diverges. But it follows from theorem 6.13 that the series n •■ i n 2 3 4 5 converges. Of course, theorem 6.13 yields no clue as to what the sum is. (It is, in fact, logg2.) Theorem 6.13 is sometimes useful, but its hypotheses are rather restrictive. A more useful theorem is the following. 6.15 Theorem (comparison test) Let 2™ =1 b n be a convergent series of posi- tive real numbers. If \a n \<b„ (« = 1,2,...) then the series 2„ = i a n converges. Proof Let e > 0. Since 2~=i b„ converges, its tail tends to zero (prop- osition 6.1 1). Hence we can find an A' such that, for any n >N, I b k <e. ft=n + l Let the sequence of partial sums of the series Z" = , a n be <s„>. Then, if n>m>N, !»«— %l = \(ai + ai + . ..+a n )-(ai + a 2 + . . . + fl m )l < |a m + il + |fl m + 2 | + . . . 4- \a„\ (triangle inequality) < I *fe<e. Thus (s„) is a Cauchy sequence and the theorem follows. Note that the hypothesis \a„\ < b n (n = 1 , 2, . . .) can be replaced by \a n \^Hb„ (n =N,N+ 1, . . .). (Why?) Note further that the hypothesis of the comparison test is not fe=i < I**. This condition is satisfied by a k = (— 1)'' and & ft = l/k 2 , but we know that 2" =1 (- 1)" diverges. Series 59 6.16 Example Prove that, if a is any positive rational number and \x\ < 1, then the series n = l converges. ft-oo/ From exercise 4.20(5) we know that n a+2 x" -+0zsn-*°°. Hence the sequence (n a+2 x n ) is bounded (theorem 4.25). It follows that, for some H, \nV\<H/n 2 (m = 1,2, . . .). Since the series S~ =1 l/« 2 converges (theorem 6.6), the result follows from the comparison test. The next two propositions are sometimes helpful. We indicate only very brief- ly how they are proved. 6.17 Proposition (ratio test) Let S„ = x a„ be a series which satisfies fln + i lim a* = I. If / > 1 , the series diverges and, if / < 1 , the series converges. If /< 1, we may take e>0 so small that /+ e< l.Then, for a sufficiently large value otN, \a„\ = a„ "n-2 °jV + 2 "n-l \a N -n\<Q + ey- N - l \a N ^\. The series S" =1 a n then converges by comparison with the geometric series 2" = i (/ + e) n . If / > 1 , a similar argument shows that the terms of 2™ = i a n do not tend to zero and so the series diverges. 6.18 Proposition (nth root test) Let ~L" =i a n be a series which satisfies lim sup \a n \ i,n m I. If / > 1 , the series diverges and, if / < 1, the series converges. If / < 1 , we may take e > so small that I + e < 1 . Then, for a sufficiently large value of TV, \a n \ <(l+ ef (n >N) (exercise 5.15(4)) and the convergence of 2"=i a n follows from the comparison test. If / > 1, e is 60 Series chosen so that / — e > 1 . Then, for some subsequence (a nk \ ' a "fe' > Q ~ e )" fe "* °° as k "* °° and so the terms of S" =1 a„ do not tend to zero. Note that if the expressions of propositions 6.17 and 6.18 diverge to + °° (instead of converging to /), then the series diverge. If / = 1 however the results yield no information about the convergence or divergence of the series at all. 6.19 Example Prove that the series T - converges for all values of x. Proof We could use the comparison test. Alternatively, if* ¥= 0, ■ as n -*■ °° (n + 1)! / n\ n + 1 and hence the series converges by the ratio test. 6.20 Absolute and conditional convergence A series 2~ =1 a„ is said to converge absolutely if the series 2"=i \a n \ converges. A series which converges but does not converge absolutely is said to be conditionally convergent. 6.21 Theorem Every absolutely convergent series is convergent. Proof Simply take b n = \a n \ in the comparison test. 6.22 Examples (i) Let a n = '• — (n =1,2,...). Then „-i n =, n 2 3 4 S ^1 111 Series 61 The first of these series converges (example 6.14). The second diverges (theorem 6.5). We conclude that the series (-1) n-l is conditionally convergent. (ii) On the other hand the series I*n=I (-1)" n = l n = l fl is absolutely convergent because I M = I k n=\ n=l n which converges. It should be noted that the comparison test, the ratio test and the nth root test all demonstrate absolute convergence. The only criterion we have given which can establish the convergence of a series which is only conditionally con- vergent is theorem 6.13 6.23 Manipulations with series Series are 'infinite sums'. It would therefore be optimistic to expect to be able to manipulate them just like 'finite sums'. Indeed, only absolutely con- vergent series may be freely manipulated. If one tries to obtain results by manipulating divergent or conditionally convergent series, only nonsense can be expected in general. 6.24 Example Consider the following argument. = + + + .. . = (1-1) + (1-1) + (1-1)+... = 1-1 + 1-1 + 1-1 + .. . = 1 +(-1 + l) + (- 1 + l) + (-l + 1) + = 1 +0 + + 0+ ... = 1. 62 Series The error is not hard to find. The series 1—1 + 1 — 1 + 1—1 divergent. (Its terms do not tend to zero.) ... is 6.25 Example We know that the series 1 2^3 4^5 6^7 •■• converges conditionally. Denote its sum by s. We rearrange the order in which the terms of the series appear and obtain i _1_i4.1_1_i4.1_1_i4.l_ 1 2 4^3 6 8^5 T5 15^7 ••• Let the Mth partial sum of this series be t n . Then 1 1 1 1 1 1 2m -1 1 In - 1 An - 2 -|I + i+... 2 6 _1_ 4m 1 4m -2 1 1 - - + -+. . . + — \4 8 An 1 1 1 +-+ . . . + 3 2/1-1 1 + 3 + + 2m -1 1/11 1 - + -+ . . . + — 2 \2 4 In 1 / 1 I = - 1 -- + +. .. 2 \ 2 3 4 2m - 1 2m \s as n Notice that the finite sum for t 3n can validly be rearranged in any way we like. Since t 3n + 1 — t 3n •* as n -*■ °° and r 3n+2 — r 3n ->0 aSM -*«»,it follows that the rearranged series converges to \s which is not the same as the sum s of our original series (because s ¥> 0). 6.26 Exercise (1) Using partial fractions, prove that f 3n ~ 2 -1, M m(m + 1)(m + 2) (2) If <a„> and <b n ) are two sequences of positive terms and Series 63 ► / as m -*■ °° b n where / <¥• 0, prove that the series £ a„ and £ b„ either both converge or both diverge. Discuss the convergence or divergence of the series (3) Suppose that <a„> is a decreasing sequence of positive terms such that _„ = ! a n converges. Prove that na„ -* as n -» °°. [ Hint: consider %*i+*»+a + • .. + «&»•] (4) Let </•„> denote the rational numbers from (0, 1) arranged in the sequence whose first few terms are \, 3, §, 4,4,4,... Prove that the series _/« diverges. (5) Determine whether or not the following series converge or diverge. ®£_? n = i(2n)! ft Cm' - ) 2 (iii) _ M + (iv) Z n = n = l \n + \j ( _ ir -. 4" (v) _. -{V(«+0-V«} n = l M n = l v« (6) If the sum of the conditionally convergent series I-i + 1-i + l-... iss, prove that the sum of the rearranged series 1+1-1 + 1 + 1-J + J + X-1 + ... is Is. Hint:r 3n = 1+ J _ 2 + '" 1 + 4n — 3 4m - 1 2m 7 FUNCTIONS 7.1 Notation A function f from a set A to a set fl (write /: A-*B) defines a rule which assigns to each x£ A a unique element y G.B. The element y is called the image of the element x and we writer = f(x). When A and B are sets of real numbers we can draw the graph of the function as in the diagram below. The defining property of a function ensures that each vertical line drawn through a point of A cuts the graph in one and only one place. /(-*) If / is a function from A to B and S C A , we say that / is defined on the set S. The largest set on which / is defined is, of course, the set A . We call A the domain of/. For example, a sequence is a function whose domain is the set M of natural numbers. /»f« » / \ 4 — w&$ ) 64 Functions 65 If / is defined on a set S, we use the notation AS) = {f(x):xeS} and say that f(S) is the image of the set S under the function /. The set f(A) is called the range off. Note that f(A) need not be the whole offi. 7.2 Example Consider the equation y = x 2 . This defines a function from U to itself. For each j£R there exists a unique y€U which satisfies the rule y — x 2 . Observe that, in the diagram below, each vertical line cuts the graph in one and only one place. The domain of this function is R. The range is [0, °°). The image, for example, of the set [-2, 1] is [0, 4]. (Why?) 7.3 Example Consider the equation y 2 = x. This does not define a func- tion from R to itself. In the diagram below, the vertical line drawn through the point x does not meet the graph at all, and hence there is no value of y associ- ated with x . 7.4 Example Again consider y 2 = x. Does this define a function from [0, °°) to R? Again the answer is no. This time it is certainly true that every 66 Functions vertical line drawn through a point of [0, °°) meets the graph (see § 1 .9). But all but one of these vertical lines meets the graph in two points. Thus, in the dia- gram, there is not a unique value of y associated with Xj. 7.5 Example Again consider y 2 — x. This does define a function from [0,~)to[0,°°). 10.°°) Since y must be in [0, °°) we omit from the diagram the part of our previous graphs which lies below the *-axis. Then every vertical line drawn through a point of [0, °°) meets the graph in one and only one point (see § 1 .9). Thus, given any x£ [0, °°), there is a unique y G [0, °°) which satisfies y 2 = x. Thus a function /is defined from [0, °°) to [0, °°). Recalling the content of § 1.9, we observe that, for each x > 0, 7.6 Polynomial and rational functions If a ,a u a 2 , ■ ■ . a n are all real numbers, then the equation y = a + a t x + a 2 x 2 + . . . + a n x" defines a function from R to itself. Any value of x which is substituted on the right hand side generates a unique corresponding value of y. If a n # 0, we call Functions 67 this function a polynomial of degree n. A polynomial of degree is called a constant. Suppose that P and Q are polynomial functions. Let S denote the set R with all the values of x for which Q(x) = removed. (If Q is of degree m, it follows from exercise 3.1 1(3) that there can be at most m such values.) Then the equa- tion P(x) y = <m defines a function from S to U. Such a function is called a rational function. 7. 7 Example The function from U to itself defined by the equation y =x 3 — 3x 2 + 2x is called a polynomial function of degree 3 (or, more loosely, a 'cubic polynomial')- Its graph is sketched below. = v'-ix 1 + 2.Y 7.8 Example Let S be the set R with 2 and - 2 removed. Then the equa- tion y = x z +4 x 2 -4 (x*±2) defines a function from S to R. Its graph is sketched below. .v ! +4 68 Functions 7.9 Combining functions We begin with some almost obvious notation. If S C IR and / and g are two functions from S to U , then we define the function /+ g to be that func- tion from S to R which satisfies (f+g)M = f(x)+g(x) (xes). Similarly, if X is any real number, we define X/ to be the function from S to IR which satisfies (x/)(x) = x/(x) (xes). Again, we define the functions fg and f/g by (.fg)(x) = f(x).g(x) (xes) (f/g){x) = f(x)lg(x) (xGS). For the latter definition to make sense, of course, it is essential thatg(x) =h for all x&S. A somewhat less trivial way of combining functions is to employ the oper- ation of composition. Let 5 and T be subsets of R and suppose that#: S-*T and /: T-* R . Then we define the composite Junction f°g:S->- R by /°*(*) = /(*(*)) (*es). Sometimes fa g is called a 'function of a function'. 7.10 Example Let/: be defined by f<*)-$ + \ (* SR) and let g: R -*■ R be defined by *(*) = * 3 - Then fog: R -+ R is given by the formula fogQc) = f(g(x)) = \g(x)} 2 + 1 x 6 X 6 -1 + r 7.11 Inverse functions Suppose that A and B are sets and that / is a function from A to B. This means that each element a S A has a unique image 6 = /(a) S 5. We say that /"' is the inverse function to / if f" 1 is a function from B toA which has the property that x = f~\y) if and only if y = f(x). Not all functions have inverse functions. In fact, it is clear that a function Functions 69 f : A -*■ B has an inverse function /"' : B -> A if and only if each b £ B is the image of a unique a^A. (Otherwise /"' could not be a function). A function which has this property is said to be a 1 : 1 correspondence between A and B. In geometric terms, a function / : A ->B is a 1 : 1 correspondence between A and B (and hence has an inverse function f~ l : £ -* A) if and only if each vertical line through A meets the graph of / in one and only one point (which makes / a function) and each horizontal line through B meets the graph of / in one and only one point (which makes f' 1 a function). 6= /(a) a =/■'») 7. /2 Example Let /: (1 , ») -* (0, 1 ) be defined by x-l /(*) = x+ 1 (* > 1). (0,1) (I.-) It seems clear from the diagram that / is a 1 : 1 correspondence between (1, ■») and (0, 1). Thus it has an inverse function. To prove this we must show that, given any y satisfying < y < 1 , there is a unique x > 1 which satisfies x-l y = (1) This is easily accomplished by solving (1) for*. We obtain y(x+ 1) = x-\ 70 Functions xiy-1) = -\-y 1+2 1-y- x = (2) Thus, if y satisfies <y < 1, there is a unique x > 1 which satisfies (1) — namely, that given by (2). We have established the existence of an inverse func- tion /"' : (0, 1) -* (1 , °°). It is given by the formula f\y) = i+y l-y Cve(o,i)). (!.«)< (O.n It is instructive to observe how the graph of x =f 1 (y) is related to that of y=m- 7.13 Bounded functions Let / be defined on S. We say that / is bounded above on 5 by the upper bound H if and only if, for any x E S, m < h. This is the same as saying that the set f{S) = {f(x)-.xes} is bounded above by H. If / is bounded above on 5, then it follows from the continuum property that it has a smallest upper bound (or supremum) on S. Suppose that B = sup /(*) = sup f(S). xBS Functions 71 It may or may not be true that, for some % G S, /(£) = B. If such a value of % does exist, we say that B is the maximum of / on the set S and that this maxi- mum is attained at the point £. /attains a maximum of B at the point £ on the set S. Similar remarks apply to lower bounds and minima. If a function /is both bounded above and below on the set S, then we simply say that / is bounded on the set S. From proposition 2.3 it follows that a function / is bounded on a set 5 if and only if, for some K, it is true that, for any x G S, l/(x)l < K. 7.14 Example Let /: (0, °°) -* R be defined by m-\ (*>o). This function is unbounded above on (0,1]. It is, however, bounded below on (0, 1 ] and attains a minimum of 1 at the point x = 1 . ymfflk 72 Functions 7.15 Example Let /: R ■* U be defined by f(x) = x. This function is bounded above on both the sets (0, 1) and fO, 1] and in both cases its supremum is 1 . But / has no maximum on the set (0. 1 ). There is no £ satisfying < | < 1 for which /(£) = 1 . On the other hand, / attains a maximum of 1 at the point x= 1 on the set [0,1]. °^fy- '//7///////////A (0,1) 7.16 Exercise (1) Draw a diagram illustrating the set of all (x, y) such that [5 if x > I 2 if x < 1. y = Explain why this is a graph of a function from R to itself. What is the range of this function? What is the image of the set [1, 2] under this function? (2) Draw a diagram illustrating the equation \x\ + \y\ = 1. [Hint: consider each quadrant separately.] Explain why (i) the equation does not define a function from U to itself; (ii) the equation does not define a function from [— 1 , 1] to itself; (hi) the equation does define a function from [—1, 1] to [0, 1]. (3) Let /: [0, 1] -» [0, 1] be defined by /(*) 1 -x (0<*<1) 1 +x and let g: [0, 1] -*■ [0, 1] be defined by g(x) = 4x(l -x) (0<x<l). Find formulae for fa g and go f and hence show that these func- tions are not the same. Show that /"' exists but that g' 1 does not exist. Find a formula for r 1 . (4) The formula y = x 2 may be employed to define a function /: U •* R . Functions 73 Explain why /has no inverse function. If, instead, we use the formula y = x 2 to define a function g: [0, °°) -> [0, °°), show that g' 1 : [0, °°] -> [0, °°) exists and that g' , (y) = \ / y (y>o). (5) A function g: A -*■ B is a 1 : 1 correspondence between A and B. Prove that (0* _1 ° *(*) = * (xGA) (ii)gog- 1 (y)=y (y&B). To what do these formulae reduce when g is as in question 4? (6) Let / and g be bounded above on S. Let c be a constant. Prove that (i) sup {/(x) + c) = c + sup f(x). iGS xGS (ii) sup {/(x) + g(x)} < sup f(x) + sup g(x). xSS xes Give an example to show that equality need not hold in (ii). 8 1 LIMITS OF FUNCTIONS Limits of functions 75 Given any e > 0, we can find a 5 > such that |/(;t)-/|<e provided that a < x < a + 5 . 8.1 Limits from the left Suppose that /is defined on an interval (a, b). We say that/(x) tends (or converges) to a limit / as * tends to b from the left and write f(x) -*■ I as x •* b — or, alternatively, lim f(x) =1 x -* b- if the following criterion is satisfied. Given any e > 0, we can find i 1 6 > such that \m -/|<e provided that b - -8<x <b. The number \f(x) — 1\ is the distance between /(x) and /. We can think of it as the error in approximating to / by f(x). The definition of the statement fix) -* / as x -* b — then amounts to the assertion that we can make the error in approximating to / by f(x) as small as we like by taking x sufficiently close to b on the left. 8.2 Limits from the right Suppose again that /is defined on an interval (a, b). We say that/(X) tends (or converges) to a limit / as x tends to a from the right and write f(x) ■* / as x -> a +, or, alternatively, lim f{x) = I x -* a+ if the following criterion is satisfied. 74 8.3 f{x) •* I as *->•§ Suppose that /is defined on an interval (a, b) except possibly for some point % G (a, b). We say that fix) tends (or converges) to a limit / as x tends to % and write f(pc) -*lasx-+% or, alternatively, if the following criterion is satisfied. Given any e > 0, we can find a 5 > such that \fix)-l\'<e provided that < \x - -IKS. If a and are real numbers, it is often useful to note that the inequality \a\ < (3 is equivalent to — < a < (3, or, what is the same thing, - (3 < — a < (see exercise 1.20(1)). Thus, in the definitions above, the condition |/(x) — /| < e can be replaced throughout by — e < fix) — / < e or, alternatively, by — e < / —fix) < e. Similarly, the condition < \x — %\ < 5 in the last definition is equivalent to the assertion -S<j:-?<6andx^|. Thus to say that < \x - 1| < 5 is to say that x satisfies one of the two inequalities £— 8<x<%or%<x<% + 8. With the help of the last remark it is easy to prove the following result. 8.4 Proposition Let /be defined on an interval (a, b) except possibly at a point % e (a, b). Then fix) -> / as x -> % if and only if fix) -* I as x -*■ % — and /(x)->/ as *-*£+. 76 Limits of functions 8.5 Example Let /be the function from R to itself defined by m = 1 -x (x<\) ,2x (x>\). Then (i) lim f(x) = (ii) lim f(x) = 2. Note that it follows from proposition 8.4 that ton f(x) does not exist even though /(l) is perfectly well defined (/(l) = 0). Proof (i) f(x) •* as x -*■ 1 — . Given any e > we must show how to find a 5 > such that |/(x)-0|<e provided that 1 — 5 <x < 1 . Since we are only concerned with values of x satisfying x < 1 , we can replace f(x) by 1 -*. The condition \f(x) - 0|< e then becomes 1 1 — x \ < e which is equivalent to — e <x — 1 < e. Adding 1 throughout, we see that | f{x) - |< e is the same as 1 - e < x < 1 + e. The problem is now reduced to the following. Given any e >. 0, find a S > such that 1 -e<x<\ +e provided that 1 — 6 < x < 1 . Obviously, the choice 6 = e suffices to make this true and this completes the proof that f(x) -*■ as x -> 1 — . (ii) f(x) -*■ 2 as x •* 1 +. Given any e > 0, we must show how to find a 5 > such that l/W-2|<e provided that 1 < x < 1 + 6. Since we are only concerned with values of x satisfying x > 1 , we can replace f(x) by 2x. The condition \f(x) - 2|< e then becomes \2x - 2| < e which is equivalent to — e < 2(x — 1) < e. Thus l/(x) - 2|< e is the same as 1 - ^e < x < 1 + |e. The problem is now reduced to the following. Given any e > 0, find a 5 > such that 1 -|e<*<l +\e provided that 1< x < 1 + 6. Limits of functions 77 Obviously, the choice 5 = ^e suffices to make this true and this completes the proof that f(x) -* 2 as x ■* 1 +. 8.6 Continuity at a point In the definition of/(x) -> / as x ■* % given in §8.3, consideration of what happens when x actually equals % was carefully excluded. When taking limits we are only interested in the behaviour of/(x) asx tends to \ not in what happens when x equals %. Thus it is quite possible for/(x) to tend to a limit / as* tends to % even though f(x) is not defined at the point % (see exercise 8.15(3) below). And, even if f(X) ' s defined, it is not necessarily true that /=/(?) (see exercise 8.15(2) below). Having said this, we can now turn to the definition of continuity at a point. Suppose that /is defined on an interval (a, b) and % e (a, b). Then we say that /is continuous at the point % if and only if /(*)-/(?) as *->£. Roughly speaking, to say that /is continuous at the point % means that the graph of/ does not have a 'break' at the point ff. If /is defined on an interval (a, b] and f(x) •* f(b) »x-*b— »we say that /is continuous on the left at the point b. If /is defined on an interval [a, b) and f(x) -*f(a) as x -> a +, then we say that /is continuous on the right at the point a. f continuous at j-. f(b)---/-\-z- / continuous on the left at b. /(a)-- /continuouson the right atfl. 8. 7 Example For any real numbers a and 0, the function /: defined by /(*) = ax + is continuous at every real number |, Proof Assume a # 0. (If a = 0, the proof is even easier.) Let e > be given. Choose 5 = e/|a|. Then, provided that \x — %\ < 5 , I/(X)-/(!)I = N*-f)l = |a|.|x-$|<|a|.8 = e. 78 Limits of Junctions Given any e > 0, we have found a 5 > such that |/(x) — /(£)| < e provided that \x — 1| < 5 . Hence f(x) -> /(£) asx -»g and so /is continuous at the point g. Note that we showed that | /(.x) — /(£) I < e 'provided that | x — g i < 5' instead of 'provided that 0<|.x — g|<5as appears in the definition of f(x) -* / as x -»• g- This is because, when x = g, it is automatically true that | /(jc) — /(g) I < e because then | f(x) — /(g) | = 0. It is only when we are considering a limit / =£/(g) that it is important to exclude the possibility that x = g. 8.8 Connexion with convergent sequences 8.9 Theorem Let /be defined on (a, b) except possibly for g € (a, b). Then f(x) •* I as x •* g if and only if, for each sequence Cx n > of points of (a, b) such that x n =£ | (n = 1 , 2, . . .) and *„ -> g as n •* °°, it is true that f(x n ) -*■ I as n -*■ °°. Proof (i) Suppose first that f(x) ■* I as x •* g. Let e > be given. Then we can find a 5 > such that \f(x) — l{<e provided that < |x — g| < 6 . Now suppose that <x n > is a sequence of points of (a, b) such that x n ¥= g (n = 1, 2, . . .) and x n ■+ 1 as « -> °°. Since 5 > we can find an iV such that for any n>N, \x n - %\< 5. But x n ¥= g (n = 1 , 2, . . .) and so < \x n - g|<5. But this implies that l/W-'Ke. Given any e > 0, we have found an N such that, for any n>N, |/(x„) - /|< e. Thus f(x„) -* / as « -* °°. (ii) Now suppose that, for each sequence (x n ) of points of (a, b) such that x n & % (» = 1 > 2, . . .) and jc„ -*■ g as « -* oo > it is true that f(x n ) •* I as n -* °°. We now suppose that it is not true that/(x) •* / as x -» g and seek a contra- diction. If it is not true that /(at) ■+ / as x ■* g then, for some e > 0, it must be true that for each value of 5 > we can find an x satisfying < |.x — g| < 6 such that \f(x)-l\>e. In particular, if S = 1/ra, then we can find an x n satisfying < \x n — g| < 1/n such that \f(x n )-H>e. But then U„) is a sequence of points of (a, b) such that x n ^ | (« = 1, 2, . . .) and x n -> I as n -*■ °° but for which it is not true that f(x„) -*■ I as n -* °°. This contradicts our assumption above. Similar results to theorem 8.9 hold for convergence from the left or right. For example, fix) -> / as x -» Z> — if and only if, for each sequence Cx n > of points of (a, 6) such that x n ■+ 6 as « -» °°, it is true that /(x n ) ->■ / as n -*■ °°. Limits of functions 79 8.10 Example We show how theorem 8.9 can be used to prove that the function /: R -* R defined by f(x) = ax + (3 is continuous at every point g (see example 8.7). Let (jc n ) be any sequence of real numbers such that x n ■* g as n ■* <*>. By proposition 4.8, ax n + ■* «| + P as n -» °°. From theorem 8.9 it follows that ax + ■* ag + p as x -* g and this concludes the proof. y=/te>=«X+/ 8.11 Properties of limits The following propositions are analogues of the combination theorem and the sandwich theorem of chapter 4. They are easily deduced from these results with the help of theorem 8.9. 8.12 Proposition Let /and g be defined on an interval (a, b) except possibly at g G (a, b). Suppose that f(x) ■* I as x ■* g and g(x) ■* m as x -* g and suppose that X and p are any real numbers. Then (i) \f(x) + pg(x) ^\l + pmasx^H (iii) f(x)lg(x) -*■ l/m as x -> g (provided m =£ 0). An important consequence of proposition 8.12 is the following result which we quote as a theorem. 8.13 Theorem A polynomial is continuous at every point. A rational func- tion is continuous at every point at which it is defined. Proof Example 8.7 shows that x -» g as x •*■ g. Repeated application of proposition 8.12(ii) then shows that x h -> g* as x •* g for each keN. Hence, if P(x) = a n x n + a n - 1 x"- 1 + . .. + a,x + a , then a repeated application of proposition 8.1 2(i) yields P(x)^P(& asx -> g, i.e. Pis continuous at g. To show that a rational function is continuous wherever it is defined we appeal to proposition 8.12(iii). 80 Limits of functions 8.14 Proposition (sandwich theorem) Let f,g and /; be defined on (a, b) except possibly at | e (a, b). Suppose that g(x) -*• / as x •*■ f , h(x) •* I as x -* % and that g(x)<f(x)<h(x) except possibly when x = %. Then/(x) -*■ I as x -*■ £. Note The analogues of propositions 8.12 and 8.14 for convergence from the left or right are both true. 15 Exercise i (1 ) Calculate the following limits ix 2 + 4) (i) lim (-= — - (ii) lim x-*i [x 2 — 4 J '*-*o (2) Let/: R -► R be defined by 3-x (*>1) m = 1 (x = 1) 2x (x<l). ,73 + 5x* 2 + 9 3x 2 Show that /(at) -*■ 2 as x -> 1 - and f(x) -* 2 as x -> 1 + using only the definitions of §8.1 and §8.2. Deduce that the limit lim f{x) X -* 1 exists but is not equal to/(l). Draw a graph. (3) Let /be defined for all x except x = by ».fl±£=I <^0). Prove that f(x) -» 2 asx -» even though /(0) is noj defined. (4) Use the definition to show that |x — £| -* as x -> £. A function /is defined on an open interval which contains % and, for each x in this interval, /(x) lies between £ and x. Prove that/(x) -> £ as x -*■ %. (5) Let n £ N . Prove that the function / : [0, °°) -» R defined by f(x) = x Vn (x>0) is continuous at each £ > and continuous on the right at 0. [Hint: Use the sandwich theorem and exercise 3.1 1(2) to prove that/is con- tinuous at £ > 0. For continuity on the right at 0, appeal to the defi- nition of f(x) -*■ as x -* +.] (6) Suppose that f(x) ■* I as x -> £. If / > 0, show that, for some h > 0, Limits of functions 81 f(x)>0 provided that £ - /; < x < £ + /i and x # £. What happens if (i) / < 0, (ii) I - 0? 8.16 Limits of composite functions Suppose that f(y) -* / asy -* 7? and that g{x) ■* 77 as x -» £. It is tempting to write y = g(x) and conclude that But some care is necessary. Consider, for example, the functions /: R -> R and g: R -*■ R defined by m = 3 = 1) 2 0*1)' g(x) = 1. Then /(j>)-»- 2 as y -* 1 andgXx)-* 1 as x -+0. But it is not true that f(g(x))->- 2 as x -* because, for all values of x,f(g(x)) =/(l) = 3. & 7 7 Theorem Suppose that f(y) -* I as y -» tj and that £(x) ->■ 77 as x ■* |. Then either of the two conditions below is sufficient to ensure that /&(*))-»/ as *-»•$. (i)/is continuous at 77 (i.e. / = /(t?)). (ii) For some open interval / containing £, it is true that g(x) # 77 for any x S 7, except possibly x = £. /Voo/ Let e > be given. Since f(y) -> / as y •*■ tj, we can find a A > such that I/O) - /|< e provided that < \y - 77I < A. Writing v =g(x), we obtain |/fe(x))-/Ke (0 provided that < \g(x) - 7?|< A (2) But £(x) -* 7? as x -* £ and A > 0. Hence we can find a 6 > such that \g(x)-v\<A ( 3 ) provided that 0<|.x-£|<5. ( 4 ) We would now like to say that (4) implies (3), which implies (2), which in turn, implies (1). This would complete the proof. But notice that (3) does not 82 Limits of functions imply (2) in general. We therefore need to use one of the hypotheses (i) or (ii) of the theorem. If we assume that/is continuous at 77, then / = /(rj) and so \f(y) —l\<e even when y = r\ (see the remarks at the end of example 8.7). Thus, in this case, we may replace condition (2) by condition (3) and the argument described above goes through. If instead we assume hypothesis (ii), then we can be sure that£(je) ^17 pro- vided that x satisfies (4) for a sufficiently small value of 5 >0. But then con- dition (3) can be replaced by condition (2) and the argument goes through again. 8.18 Divergence We write down two sample definitions. The reader will have little diffi- culty in supplying the definitions in other cases. We say that/(x)->--l- °°as*-> £+ if, given any /f > 0, we can find a 5 >0 such that provided that | <x < g + 6. We say that f(x) ■* / as x -*■ — °° if, given any e > 0, we can find an X such that |/(x)-/|<e provided that x < X. f{x)—+->*sx^'i + f(x)-'l as x-*- 8.19 Example Prove that x l -* + °°asx->0+. Proof Let H > be given. We have to find a 5 > such that 1 ~>H x provided that <x < 8 . Obviously the choice 8 = H' 1 satisfies the requirement. Limits of functions 83 120 Exercise (1) Prove the following (i) x' 1 -*■ — °°asx->0 — (ii) x 7, -* + °° as x -* + °°. (2) Suppose that f(y) ■* I as y ■* + « a nd g(x) ■* + °° as x ■* + «. Prove that/(g(jc)) -»■ / as x ■* + °°. Explain why the difficulties encountered in theorem 8.17 do not occur in this case. (3) Suppose that f(y) -+ 1 as y ■* +. Prove that fix' 1 ) -+l asx-> + °°. (4) Suppose that /is defined on (a, b) except possibly for | G (a, b). Let (x n ) be a sequence of points of (a, b) such that x n i= % (n = 1 , 2, . . .) and x n -*■ % as n -* °°. Let (y„) be a sequence with the same properties. If / ¥= m and /(*„) -*■ / as n ■* °° but /(y n ) -*■ m as M -* °°, show that does not exist. [Hint: theorem 8.9] (5) Let/: U^U be defined by (1 (x rational) (x irrational). Show that lim f(x) x -* does not exist. (6) If /is defined as in the previous question, prove that iim {*/(*)} = 0. I-»0 CONTINUITY 9.1 Continuity on an interval A function / is continuous at a point % if and only if f(x) -*/(£) as x •* | (see §8.6). Geometrically, this means that the graph of / does not have a 'break' or 'jump' above the point £. To say that / is continuous on an interval / should mean that its graph is unbroken above the interval /, i.e. that we can draw the part of the graph which lies above /without lifting our pencil from the paper. A precise mathematical definition is required. This depends on the type of interval in question. A function / is continuous on an open interval / if and only if it is continuous at each point of/. Where an endpoint of an interval belongs to the interval, a slightly more com- plicated definition is required. The most important case is that of a compact interval [a,b]. A function / is continuous on a compact interval [a, b] if and only if it is continuous at each point of (a, b) and continuous on the right at a and on the left at b. /continuous on ia.b) -^ mmt&i /continuous on \a,b) 84 Continuity 85 Note that, in both diagrams, the graph of /is unbroken above the interval in question. In general, a function / is continuous on an interval / provided that: (i) it is continuous at each point of/ which is not an endpoint; (ii) it is continuous on the right at the left hand endpoint of/, // this exists and belongs to /; (iii) it is continuous on the left at the right hand endpoint of/, //"this exists and belongs to /. 9.2 Examples (i) All polynomials are continuous on every interval. All rational func- tions are continuous on any interval not containing a zero of the denominator (theorem 8.13). (ii) Let /: (0, ■»)-»• R be defined by f(x) = 1/jc. Then / is continuous on the interval (0, °°) because it is continuous at each point % satisfying £ > 0. (iii) Let /: U -+R be defined by (x 2 (jc<1) \ 3/2 (*>1). Then / is continuous at every point except x = 1 . But / is continuous on the left at x = 1 and thus /is continuous on the interval [0,1]. It is not continuous on [1,2]. 9.3 Proposition Let / be defined on an interval /. Then / is continuous on / if and only if, given any x G / and any e > 0, we can find a 5 > such that \f(x)-f(y)\<e provided that y € /and satisfies \x —y\ <8. 86 Continuity 9.4 Proposition Let X and n be real numbers and suppose that the func- tions /and g are continuous on an interval /. Then so are the functions (i)X/+W? PDA (iii) fig (provided g(x) ¥• for any x e /). 9.5 Proposition Let g : / -*■ J be continuous on the interval / and let /: J-*R be continuous on the interval /. Then / o g is continuous on /. 9.6 Proposition Let /be continuous on the interval /. If £6/ and (x n ) is a sequence of points of / such that x n -* if as n -*■ °°, then These four propositions are easily proved. The first is proved by an appeal to the limit definitions of § §8.1, 8.2 and 8.3. The second follows from proposition 8.2, the third from theorem 8.17 and the fourth from theorem 8.9. The last of these results can be remembered in the form lim /(*„) = /(lim x n ) i.e. for a continuous function the limit and function symbols can be validly exchanged. (We say that the symbols 'commute'.) 9.7 Continuity property We now come to a theorem which may seem so obvious as to be hardly worth mentioning. It is, however, of a fundamental importance comparable to that of the Continuum Property of §2.2 and will serve as a foundation stone in our development of the calculus in the next few chapters. 9.8 Theorem (continuity property) Let /be continuous on a compact interval [a, b] . Then the image of [a, b] under /is also a compact interval. \c.d \a.b\ Continuity 87 We split the content of the continuity property into three lesser theorems and prove these separately. Given that /is continuous on 1 = [a, b] , it follows from theorem 9.9 that /=/(/) is an interval. From theorem 9.1 1 it follows that the interval /is bounded and from theorem 9.12 that J includes its endpoints. 9.9 Theorem Let /be continuous on an interval /. Then the image of / under/is also an interval. Proof To show that J =f(I)={ f(x) : x € / } is an interval , we need to show that, if y x andy 2 are elements of J and j>, < X <y z , then X EJ (see §2.9). Since y i £ J and y 2 e J, the subsets of / defined by S = {x:f(x)<\}; T = {x:f(x)>\} are non-empty. Also, every point of the interval /belongs to one or other of these sets. It follows that a point of one of the sets is at zero distance from the other (exercise 2.13 (6)). Suppose that s E 5 is at zero distance from T. Then a sequence < t n ) of points of T can be found such that t n -*■ s as n -* °° (exercise 4.29(6)). Since/is continuous on / it follows fhat/(r„) -»/(s) as n -* °° (pro- position 9.6). But f(t n )>\ (n = 1,2,...) and therefore /(s) > X (theorem 4.23). We already know that f(s) < X and so it follows that /(s) = X. Hence X EJ. A similar argument applies if a point of T is at zero distance from S. 9.10 Corollary (intermediate value theorem) Let /be continuous on an interval / containing a and b . If X lies between f(a) and f(b), then we can find a % between a and b such that X = /(£). fib) Proof The corollary is simply a somewhat clumsy restatement of theorem 9.9. 9.11 Theorem Let /be continuous on the compact interval [a, b]. Then /is bounded on [a, 6]. Proof Suppose that /is unbounded on [a, b]. Then we can find a 88 Continuity Continuity 89 sequence (x n ) of points of [a,b] such that |/(x B )I-*- + 0O as»-+« (1) (see exercise 4.29(6)). Since [a, b] is compact, there exists a subsequence <x„ r > .which converges to a point £ £ [a, b] by the Bolzano-Weierstrass theorem (see exercise 5.21(4)). Because /is continuous on [a, b\ ,/(x„ r ) -*/(£) as r ->■ °° (pro- position 9.6). But this contradicts (1). 9.72 Theorem Let /be continuous on the compact interval [a, b]. Then/ achieves a maximum value d and a minimum value c on [a, b] . Proof We know from the previous theorem that /is bounded on [a, b]. Let <2 be the supremum of /on [a, b] and consider a sequence < X n ) of points of [a, b] such that f(x n ) -*■ d as « -* °° (see exercise 4.29(6)). Since [a, b] is com- pact, (x n ) contains a subsequence <x„ r > which converges to a point * S [a, &], Because /is continuous on [a, b], it follows that /(x„ r ) -*/(£) as/--* <*>. Hence ^•(H) = ^ and thus the supremum d is actually a maximum. A similar argument shows the infimum to be a minimum. 9.13 Example The function/: R -* R defined by 5 (x>l) /(*) = 2 (*<1) is not continuous on [0,2]. Observe that 2 = /(0) < 4 </(2) = 5 but there is no value of £ between and 2 such that /(£) = 4. Note that, for X satisfying 2<X<5, {x:f(x)<\} = {2}; {x:f{x)>\} = {5}. 9.14 Example The function/: (0, °°) ■* R defined by /(x) = 1/x is con- tinuous on the open interval (0, 1) but is not bounded on (0, 1). The function g : R ■* R defined by g(x) = x is continuous everywhere and happens to be bounded on the open interval (0, 1). But g does not attain a maximum value or a minimum value on (0, 1). 9.15 Example Show that the equation 17x 7 -19a- 5 -1 = has a solution % which satisfies — 1 < | < 0. Proof The function/: R-*R defined by /(x) = 17x 7 - 19.v 5 - 1 is a polynomial and hence is continuous everywhere. In particular it is continuous on [-1,0]. Now/(- 1) = 1 and /(0) = - 1 . Since - 1< < 1 , it follows from corollary 9.10, that/(£) = for some | between — 1 and 0. 9.16 Example Let/: [a, b] some £ G [a, ft], m) = t [a, b\ be continuous on [a, b] . Then, for Thus a continuous function from a compact interval to itself 'fixes' some point of the interval. (This is the one-dimensional version of Brouwer's fixed point theorem.) Proof The image of [a, b] is a subset of [a,b]. Thus f(a) > a and fib) < b. The function g : [a, b] •* U defined by g(x) = f(x) — x is continuous on [a, b] by proposition 9.4. But g(a)>0 and g(b) <0. By corollary 9.10, for some £ e [a, b] it is true that£(£) = 0. Thus/(£) = %. 9.17 Exercise (1) Each of the following expressions defines a function on (— 2, 2). Decide in each case whether or not the function is continuous on (a) (— 2, 2) and (b) [0, 1 ] . In each case draw a graph. (i) /(*) = 2x4-3 2x-5 (ii) fix) = \x-l\ 90 Continuity (1 (*<1) (1 (0<x<l) (iii) f(x) = { (iv) fix) - j \2 (x>l) iO (otherwise) (v) /(*) = * 2 + 4 x 2 -4 (vi) /(*) = x(x-\) (0<*<1) (otherwise) Which of these functions are bounded on (- 2, 2)1 Which attain a maximum on (— 2, 2)? Which attain a minimum on (— 2, 2)? (2) A function /is continuous on an interval / and for each rational number r £ / it is true that f(r) = r\ Prove that/(x) = x 1 for any xE.1. [Hint: proposition 9.6.] (3) Show that all polynomials of odd degree have at least one (real) root. (4) Suppose that / is continuous at every point and that f(x) -> as x -» + oo and/(jc) -*■ as x -» — °°. Prove that /attains a maximum value or a minimum value on the set U. (5) Let / be continuous on the compact interval /. Suppose that, for each x £ /, there exists a y £ / such that l/OOKil/001- Prove the existence of a % £ / for which /(£) = 0. [Hint: Bolzano- Weierstrass theorem.] (6) Let / be a closed interval and let < a < 1 . Let /: / -*■ / satisfy the inequality \ax)-f(y)\<a\x-y\ for each x £ / and y £ /. (Such a function is called a contraction map- ping). Prove that /is continuous on /. Let x, £ / and define x n+l - /(*„) (n = 1 , 2, . . .). Prove that the sequence <x n > converges and that its limit / satisfies /=/(/). [Hint: exercise 5.21(1).] 10 D IFFERENTIATION 10.1 Derivatives Suppose that /is defined on an open interval /containing the point £. Then /is said to be differentiable at the point £ if and only if the limit ,. /(*)-/(£) lim z — *-** x-% exists. If the limit exists, it is called the derivative of /at the point % and denoted by/'(£) or £>/«). For a function /which is differentiable at % we therefore have /'(I) = Hm, m-rm ■ i x — | Equivalently, we may write ffi + h)-f® fm - Hm h (1) From an intuitive point of view, to say that a function /is differentiable at % means that a tangent can be drawn to the curve y = fix) where x = £. The slope or gradient of this tangent is then equal to /'(£)• This geometric interpretation is based on the diagram below. /(A) r-/(£> + /(.v {) 91 92 Differen tiatiun In order that the line y = /(?) + l(x — |) be tangent to y = fix) where x — % t we require that the slope of the chord PQ approach the slope / of this line as x -*■ £. The slope of the chord PQ is qr _ m-m PR x-% and hence our requirement reduces to lim f^lM . , *•** X— | as in the definition of differentiability. (Note: It follows immediately from our definition that a function/is differen- tiable at ij if and only if there exists an / such that m + 1') -f® - if' = o(h) (h -* o). (2) Here o(h) denotes a quantity which tends to zero when divided by h. This form of the definition lends itself more readily to generalisation.) 10.2 Higher derivatives If /is differentiable at each point of an open interval /, we say that/is differentiable on /. In this case it is natural to define/' or Df to be the function from / to R whose value at each point x £ / is f'(x). We can then define the second derivative /"(£) or D 2 f(g) at the point % by f m = I™ /'(*)-/'(*) 1 x-% provided that the limit exists. Similarly for third derivatives and so on. 10.3 Examples (i) Let/: R->R be defined by f(x) = x 2 . Then /'(*) = Dx 2 = 2x for all values of x. Proof We have f(x + h) -f(x) (x + hf -x 2 2xh+ h 2 h h h = 2x + h -*■ 2x as h -*■ 0. (Note that we studiously ignore what happens when h = - see §8.4. It is as well that this is permissible, for the expression f(x + h)-f(x) Differentiation 93 is quite meaningless when h = 0.) (ii) Let/: (0, <*>) -* R be defined by f(x) = 1/x (x > 0). Then fXx)=D l -=-± for all values of x > 0. Proof We have f(x + h)-f(x) \_ h h 1 1 -h h \x(x + ft) I x + h x - — * -a as h -*■ 0. x(x + h) x 2 10.4 More notation We shall avoid the notation introduced in this section because, when using this notation, it is difficult to make statements with a precision adequate for our purposes. In many applications, however, it is unnecessary to maintain this level of precision. In such cases, the notation described below can be a very useful and powerful tool and it would be pedantic to deny oneself the use of it. Replace £ by x and h by 5x in formula (1) of § 1 0.1 . Then /'(*) = lim &x -• o f(x + 8x)-f(x) 8x = lim — Sx - o S* (3) If>> =f(x), the quantity 5^ =f(x + 8x) —f(x) is usually said to be the 'small change in y consequent on the small change 8x in*'. From formula (3) it is only a short step to the much used (and commonly abused) notation This notation has the drawback that it uses the symbol x ambiguously. It is used simultaneously as the point at which the derivative is to be evaluated and the variable with respect to which one is differentiating. In applications where this distinction is unimportant, the ambiguity creates no problems. But, for the material of this book, the distinction is important. Further confusion is possible when 'differentials' are introduced. The differ- ential df of a function /may be regarded as the function of two variables given by df(x;h) = /'(*)//. (4) Formula (2) can then be rewritten as 94 Differentiation f{x +h)- f{x) -dfix-h) = o in) ih ■+ 0) (5) provided that /is differentiable at*. For a fixed value of*, the equation k = dfix-h) = f\x)h (6) is the equation of a straight line. If the h and k axes are drawn as below, (6) is the equation of the tangent to the graph of the function /at the point*. A" =/'(.v)/i The reader may justifiably feel that little or nothing has been added to our understanding by the introduction of the idea of a differential, the idea being more useful when functions of several variables are being considered. Our only reason for discussing differentials is to explain the meaning of the equation dy = f'ix)dx. This is simply equation (6) with the variable h replaced by dx.and the variable k replaced by dy. Thus dx and dy should be thought of as variables. The reason one chooses this notation is to make the formula dy = -fdx dx (7) a valid one. It is false that dy denotes 'the small change in y consequent on a small change dx in *'. Formula (5) tells us that this is nearly true for small values of dx but it will only be exactly true for functions /with a straight line graph. Should it be necessary to discuss 'small changes in * and >>', the notation 5* and 8y is available and one can say that 8y is approximately equal to f'ix)bx. Even more false (if that is possible) is the idea that dy and dx are somehow 'infinitesimal quantities' obtained by allowing 5* and 8y to tend to zero. One would then regard dy dx Differen tia tion 95 as the quotient of these quantities and abandon all the apparatus of limits. This metaphysical notion is admittedly an attractive one and it is on this idea that Newton based his 'Theory of Fluxions'. Unfortunately it cannot stand up to close logical scrutiny. (Note: A common error is to confuse the terms derivative and differential. A derivative is sometimes referred to as a 'differential coefficient' (because of (7) above) but should not be called a differential.) 10.5 Properties of differentiable functions 10.6 Theorem Let /be defined on an open interval / which contains the point £. If /is differentiable at £, then /is continuous at %. Proof We have /(*)-/(£) fix) -m) = x-% (*-£W(S)-0as*^. Hence /(*) -*/(£) as x -*■ | and the proof is complete. 10. 7 Example Let /: U •* U be defined by [1 (x > 1) -1 (*<1). m = We can see immediately that /is not differentiable at the point 1 because it is not continuous there. 10.8 Example Let/: R -> R be defined by /(*) = * ix>\) x 2 ix<\). Then /is continuous at 1 but not differentiable (The graph has a 'corner'). We have Iim ; — lim ; — h->0+ h h--0+ h = 1 r /0+^)-/(l) ,. (1+/Q 2 -1 . lim ; = lim ; = 2. h-»o- h h->o- h Thus the converse of theorem 10.6 is false. 96 Differentiation 10.9 Theorem Suppose that /and g are defined on an open interval /con- taining the point %. Let X and \x be any real numbers. Then, if /and g are differentiable at £, (i) £>{X/+WT} = W + liDg (ii) D{fg) = fDg + gDf (iii) D j4= gJ^Zgte (provided *(g) * ) at the point f. /Voo/ 1 0) J (X/(? + A) + «?(! + h) - \f® - WG)} = X m+h)-m h + H g(M+m-g® («) A 1 = jf {/(i + am* + a) -/ct)r(i + h) +m)ga + h) -/«)*«)} = *ci + /o m+h)-m) h +/«) g (S+*)-g( l) h (Note that£(i; + A) ->-£•(£) as A -* because, by theorem 10.6, g is continuous at?.) m 1 ( flt+*) ftm _ /g+ft)gg)-g(s+ft)/(S) = » L m m+h)-m _ m g(.i+h)-m i fed)} 5 {*«)/'(*) -/«)*'(S)} as A -0. 70.70 Example If « is a natural number, then, for any x, Dx" = nx"-\ Proof This is obvious when n = 1 since (x + h)-x = 1 -*■ 1 as A •* 0. Differentiation 97 Now assume its truth when « = k. Thus Z>.x fe "fee*" 1 . Using theorem 1 0.9 (ii), D(x h + l ) = D(x h .x) = x h .l +x.kx k - 1 = (A+l)x fc and the result follows by induction. 10.11 Exercise (1) Prove that, for any x. D 1 1 +x' -2x (1 +xy (2) If n is a negative integer and .v 9* 0, prove that jDx" = nx n ~ 1 . (3) Let/: RHR be defined by /(*) = 2x (x>l) x 2 +1 (x< 1). Prove that /is differentiable at the point 1 and has derivative 2. Let g: R ■* R be defined by g(x) =\x\. Prove that g is not differen- tiable at the point 0. [Hint: Consider the right and left hand limits separately.] (4) A polynomial P of degree n has the property that />(£) = and />'(g) = 0. Prove that P{x) - (pc-tfQ(x) for all x, where Q(x) is a polynomial of degree n — 2. (See exercise 3.11(3).) (5) Let/: R-*R be n times differentiable at the point g. Show that the 'Taylor polynomial' P defined by pw = f® + ^rfm + ^=rVa) + • ■ ■ + (•*-£)" fCn-iy (« - 1)! -/»-»(|) has the property P*\|) = / w (g) (* = 0, 1 , 2, . . . , n - 1 ). (Note that / <fc) (l) denotes the derivative of order k at the point jf.) ^(6) Let/and,? be n times differentiable at the point %. Prove 'Leibniz's rule', i.e. D n fg = t ("W/EP-'f ;=o \/ at the point g. [Hint: Recall the proof of the binomial theorem.] 98 Differentiation 10.12 Composite functions The next theorem is usually remembered in the form dz dz dy dx dy'dx 10.13 Theorem Suppose that g is differentiable at the point x and that/is differcntiable at the point y = g(x). Then (f°g)'(x) = f'(y)g'(x). Proof Recall thatfog(x) =f(g(x)). We therefore have to consider f(g(x+h))-f(g(x)) h Write k = g(x +h) —g(x). Since g is differentiable atx, it is continuous there and so k -*■ as h -*■ 0. Suppose that k=£0. Then f(g(x + h))-f(g(x)) _ f(gix+h))-f(g(x)) g(x+h)-g(x) h g(x + h)-g(x) ' h __ f(y + k)-f(y) g(x+h)-g(x) k h It is tempting to deduce the conclusion of the theorem immediately. But care is necessary. What happens if, for some values of h, k = g(x +h)—g(x) = 0? To get round this difficulty we introduce the function (fjy + k)~f(y) F(k) - ( (* = 0). k f'(y) (**0) Since /is differentiable sty, F(k) -+f'(y) as fc ->- 0. Because F(0) =f'(y), it follows that F is continuous at the point 0. From theorem 8.1 7(i) we deduce that F(k)^f'(y)ash^0 where k = g(x + h) —g(x). For k^Ov/e have shown that f(g(x+h))-f(s(x)) g(x + h)-g(x) = F(k) . - h h But this equation is also true when k = since then both sides are zero. It follows that f(£(x + h))-f(g(x)) f'(y)g'(x) a sh^Q and this is what had to be proved. Differentiation 10.14 Example D{\ + a-"} 100 = l°0O + * 97 )".97x 96 . Proof Set/0) =y 100 andg(x) = l+x 97 . By theorem 10.13, D{\ +x 97 } 100 - DfXgfx)) = f'(g(x))g'(x) = 100fe(x))".97x 96 = 100{1 +x 97 }".97x 96 99 10.15 Exercise (1) If* >0 and «SN, prove that D{x vn \ = -x lln x-K n [Hint: exercise 3.1 1(6). Recall that the continuity of the nth root func- tion was considered in exercise 8.15(5).] (2) Use the previous question and the rule for differentiating a composite function (theorem 10.13) to show that, for any rational number r, D{x r ) = rx r ~' provided thatx>0. (3) Evaluate the following derivatives for those values of x for which they exist. (i) D{ 1 + x V2 Y s (ii) Dy/(X + V(* + V*)). ' (4) Suppose that whenx=l. Prove that/'(l) = or /(l) = 1. (5) Let/: R->R be differentiable at x. Suppose that/admits an inverse function/" 1 : U-*U which is differentiable atj =f(x). Prove that DT l (y) = l Df(x) ■ [Hint: f' 1 °f(x) = x\ . The function g: R^-R defined by g(x) = x 3 admits an inverse function #""': U-*U. Isg' 1 differentiable at the point 0? t(6) A function/: R^-R is defined by m = x (x rational) — x (x irrational). Show that/o/(x) = x for all values of x. What may be deduced from theorem 10.13? 11 MEAN VALUE THEOREMS 11.1 Local maxima and minima Let/be defined on an open interval (a, b) and let £ E {a, b). We say that /has a local maximum at £ if m<m for all values of x in some open interval / which contains %. Similarly for a local minimum. / has beat max al ?. /has local min al {. Very roughly, /has a local maximum at £ if its graph has a 'little hill' above the point £. Similarly,/ has a local minimum at | if its graph has a 'little valley' above the point %. J f/(?) is the maximum value of/ on the whole interval (a, b), then obviously /has a local maximum at £. But the converse need not be true (see the diagrams above). 11.2 Theorem Suppose that/is differentiable on (a, ft) and that £G (a, b). If /has a local maximum or minimum at |, then /'a) = o. 100 heal min Mean value theorems 101 Proof By definition /(*)-/(£) *-? /'(S)asx^?. Suppose that /'(£) > 0. From Exercise 8.15(6), it follows that for some open interval /=(£-/?,? + h) x-% >0 (1) provided that xG 1 and xl=%. Let x, be any number in the interval (£ — h, %). Then x t — % < and hence it follows from (1) that/(x,) < /(£). Thus /cannot have a local minimum at £. Let x 2 be any number in the interval (£, £ + h). Then x 2 — % > and so it follows from (1) that/(x 2 ) >/(!). Thus/cannot have a local maximum at %. /'(£)> { *i 'b A similar argument applied to —/deals with the case when /'(?) < 0. The only remaining possibility is/ (if) = 0. 1 1 .3 Stationary points A point | at which /'(£) = is called a stationary point of/. Not all stationary points give rise to a local maximum or local minimum. The reader will be familiar with the case when /'(?) = and /has a point of inflexion at % as in the diagrams below. f\i) = o 102 Mean value theorems But worse behaviour than this is possible. The diagram below illustrates a case where /'(?) = but /has no local maximum, no local minimum, and no point of inflexion at £. (See the solution to exercise 16.5(6).) 11.4 Theorem (Rolle's theorem) Suppose that /is continuous on [a, b] and differentiable on (a, b). If /(a) = f(b), then, for some % G (a, b), f'Q) = 0. /'(?) = Xa)=f(b) — Proof Since /is continuous on the compact interval [a, b],it follows from the continuity property (theorem 9.12) that/attains a maximum M at some point £i in the interval [a, b] and attains a minimum m at some point g 2 in the interval [a, b]. Suppose £, and % 2 are both endpoints of [a, b] . Because f(a) = f(b) it then follows that m = M and hence / is constant on [a, b] . But then /'({•) = for all Suppose that | t is not an endpoint of [a, b] . Then £, G (a, b) and /has a local maximum at £,. Thus, by theorem 11. 4, /'(£]) = 0. Similarly if | 2 is not an end- point of [a,b]. [Note the importance of the continuity property in this proof.] 1 1 .5 Mean value theorem U-6 Theorem {mean value theorem) Suppose that /is continuous on [a, b] and differentiable on (a, b). Then, for some % G (a, b), Mean value theorems Kb) -m 103 /'(«) = b-a In the diagram, the slope of the chord PQ is f(b)-Ka) b-a Thus, for some % G (a, b), the tangent to /at % is parallel to PQ. Proof LetFbe defined on [a, b] by F(x) = f(x) + hx where h is a constant. Then F is continuous on [a, b] and differentiable on (a, b). We choose the constant /( so that F(a) = F(b). Then and so /(«) + ha = f(b) + hb h = - m -m b-a Since F satisfies the conditions of Rolle's theorem, F '(£) = for some % G (a, b). But then F'{%) = f'm + h - and the theorem follows. 11.7 Theorem Suppose that /is continuous on [a,b] and differentiable on (a, b). lff'(x) = for each x G (a, b), then /is constant on [a, b] . Proof Let y G (a, b] . Then /satisfies the conditions of the mean value theorem on [a,y] . Hence, for some £ G (a,y), f'CO = m-m y-a But /'(£) = and so f(y) = /(a) for any y G [a, b\ . (Note: This theorem will be invaluable in chapter 13, about integration. It may seem intuitively obvious but, without the mean value theorem (which depends ultimately on the continuity property), it would be very hard to prove.) 11.8 Exercise (1) Find the stationary points of the function /: U ~* D defined by 104 Mean value theorems f(x) = x(x — l)(x — 2). Determine the maximum and minimum values of/on the interval [0, 3] . (2) Suppose that /and g are continuous on [a, b] , differentiable on (a, b) and that g'(x) =£ for any x G (a, b). Prove that, for some | S (a, b), g'm g{b)-g{a)- This is the Cauchy mean value theorem. [Hint: consider F = f+ hg where h is a constant.) (3) In addition to the hypotheses of question 2, suppose that /(a) = g(a) — 0. Deduce that b « - to as provided that the second limit exists. This is L'Hopital's rule. [Hint: exercise 8.15(4).] Use L'Hopital's rule to evaluate lira {y-y/V+y 1 )} is differentiable at every point and that [Hint: put_y = x l . (4) Suppose that /: U - /'(*) = x 2 for all x. Prove that f(x) = |ar 4- c, where c is a constant. * (5) Let /: U •* U be n times differentiable at every point. Let P be a polynomial of degree n — 1 for which mi) = pai) CF-o. !,...«) where | > £i> ■ ■ ■ , % n are a 'l distinct points. Prove that, for some |, / <n) (?) = 0. *(6) Letg-: U~*-R satisfy ^(0) =^(1) = and suppose that/: R->R is differentiable at every point and y = f(x) is a solution of the differen- tial equation dy g(x)f+y = 1. dx Prove that/(x) = 1 for any x satisfying < x < 1 . [Hint: Use Rolle's theorem to show that any open subinterval of (0, 1) contains a point £ at which /(|) = 1 . Then appeal to the continuity of /in the manner of exercise 9.17(2).] Mean value theorems 1 1 .9 Taylor's theorem Suppose that /is differentiable at the point £. Then the equation of the tangent to y = fix) at the point % is y =/© + (* -£)/'«)• 105 y=f(x) s'y = M) + U- ?)/'«) Thus, for values of x which are 'close' to £, it is reasonable to expect that /(£) "•"(•*"" £)/'(£) i s a 'good' approximation to/(x). But how large may the error in this approximation be? Suppose now that /is n — 1 times differentiable at the point %. Then the first n — 1 derivatives at the point % of the polynomial p(x) = /(?) + yr(* - tr'cu + ^ - ?) 2 /"od + • • • + ^ L T)i (x " r " /< "" 1>(l) are the same as those of/. (See exercise 10.1 1(5).) We might therefore reasonably expect that P(x) will be a 'very good' approximation to f(x) for values of x which are 'close' to £. But again, how large may the error be in this approximation? ,->' 11 1/ y = /'(.v) y = fix) V — — — — ~y ' >- i These questions are answered to some extent by the following version of Taylor's theorem. If n = 1, the theorem reduces to theorem 1 1 .6 (the mean value theorem). One may therefore regard Taylor's theorem as the Kth order mean value theorem. 11.10 Theorem (Taylor's theorem) Suppose that /is n times differentiable on an open interval / which contains the point %. Given any x G I, fix) = m) + ^ ( X - !)/*<D + i (x - oY'm + ■■■ + i («-l)! &-& , - 1 f i *-%) + E„ 106 Mean value theorems where the error B„ satisfies n\ for some value of 7? between if andx. Proof liy lies between x and g, put ^(y) = fix) -f(y) - (x -^)/'(y) - ... - ( *~_^" ) , 1 / ' ' "CO- Taylor's theorem asserts that, for some t? between .* and g, m) = ^(x-m in \n) n\ The function F has the simplifying property that Mean value theorems 107 (2) F'O) = The function (» - 0! /<»>(y) has the property that G(£) = G(x) = 0. It follows from Rolle's theorem that there exists an 77 between x and % for which = G'cn) = nq)+ irtr *TS) (n-l)! 7 w ((x-|) n rw Identity (2) follows. but/ (n) (g) # 0. In each of the following cases decide whether or not / has a local maximum or a local minimum at g. In those cases where/ has neither a local maximum nor a local minimum discuss its behaviour close to the point £. (i) / <n> (g) > and n even. (ii) / ( "%)>0andHodd. (iii)/ <n) (S)<0 and n even. (iv)/ (n) (£)<0and«odd. [Hint: exercise 8.15(6).] Discuss the behaviour of the functions given below at the point 0. (v)/(x) = x 93 (vi)g(x) = x 94 (vu)A(x) = -x 96 72.77 Exercise (1) IfnENandxE R, use Taylor's theorem to prove the binomial theorem in the form n(n-\) , (1 + *)" = 1 +nx+ K x 2 x . (2) Use Taylor's theorem with n = 3 and £ = 4 to obtain an approximation to y/5 for which the error is at most 2" 9 . (3) Let /be n times differentiable on an open interval 7 which contains the point if and suppose that D"/is continuous at the point g. You are given fm = /"© =...=/ <n " l) (£) = MONOTONE FUNCTIONS 12.1 Definitions Let /be defined on a set S. We say that f increases on the set S if and only if, for each x G S and y G S with x <y, it is true that If strict inequality always holds, we say that /is strictly increasing on the set S. Similar definitions hold for decreasing and strictly decreasing. A function which is either increasing or decreasing is called monotone. A function which is both increasing and decreasing must be a constant. 12.2 Examples The function/: K -*U defined by f(x) =x 3 is strictly increasing on R . The function/: (0, °°) ->R defined by f(x) = l/x is strictly decreasing on (0, °°). y 12.3 Limits of monotone functions We introduce the notation /($-) = lim f{x); /(?+) - lim /(*) x->$- x-*i+ provided that these limits exist. Do not confuse /(£ — ) and /(£ +) with/(£), which is the value of the function/at £f. In example 8.5, /(l — ) = = f(\)but /(l +) = 2. 108 Monotone functions 109 12.4 Theorem (i) Let /be increasing and bounded above on (a, b) with smallest upper bound L. Then f{x) -* L as x -* b — . (ii) Let /be increasing and bounded below on (a, b) with largest lower bound /. Then f{x) -> / as x -*a +. Proof We only prove (i). Let e > be given. We have to find a 5 > such that, given any x satisfying b — 8 < x < b. i.e. l/<X>-/.Ke L-e<f(x)<L + e. The inequality f(x) < L + 6 is automatically satisfied because L is an upper bound for/on (a, b). Since L - e is not an upper bound for/on (a, b), there exists a y G (a, b) such that /( y) >L-e. But /increases on (a, b). Therefore, for any x satisfying y<x<b, L-e<f(y)<f(x). The choice 5 = b —y then completes the proof. 12.5 Corollary Let /be increasing on (a, b). If £ G (a, b), then /(£ — ) and /(£ +) both exist and /(*) </(i -) </(?) </(? +) </(y) provided that a <x<%<y<b. /(£+> Ptoo/ The function /is bounded above on the interval (a, %) by /(£). By theorem 12.4, the smallest upper bound is/(£ -). It follows that, for any 110 Monotone functions fix) </(*-) </(*). A similar argument for the interval (£, 6) yields the other inequalities. 12.6 Differentiable monotone functions 12. 7 Theorem Suppose thai /is continuous on fa, b] and differentiable on (a, b). (i) If f'{x) > for each x G (a, b), then /is increasing on [a, b] . If fix) > for each x G (a, b), then /is strictly increasing on [a, b] . (ii) If f'(x) < for each x G (a, b), then /is decreasing on [a,b]. If f\x) < for each xE(a,b), then/is strictly decreasing on [a,b]. Proof Let c and d be a/ty numbers in [a, b] which satisfy c <d. Then /satisfies the conditions of the mean value theorem on [c,d\ and hence, for some if G(c,rf)> fm = f(d)-m d-c ' If /'(*) > for each x £ (a, b), then /'(£) > and hence f(d) >f(c). Thus / increases on [a, b]. If fix) > for each x G (a, b), then /'(£) > and hence fid) >/(c). Thus / is strictly increasing on [a, b]. Similarly, for the other cases. 12.8 Example Consider the function/: R-*-|R defined by f(x) =x(l —x). We have f\x) =\-2x. Hence f(x) > when x < \ and /'(*) < when x > \ It follows that / increases on (— °°, j] and decreases on \\, <»). 12.9 Inverse functions A strictly increasing function /defines a 1 :1 correspondence between Monotone functions 111 its domain / and its range /. Hence / always has an inverse function / "' and this is strictly increasing on J. (Proof?) /-■?/— / In the case when / is an interval and / is continuous on /, then J = /(/) is also an interval (theorem 9 .9) . Observe that / _1 : J -* / is continuous on /. This is a simple consequence of corollary 12.5. (If/" 1 had a discontinuity at % G/, then we could find a X G / for which /(X) would be undefined.) / ' discontinuous at if r 1 (£-)<x </-'(£+) /(X) undefined The next theorem is usually remembered in the form dx = (dy\ dy \dx) although this notation begs a lot of questions. 12.10 Theorem Let / and / be intervals and let 7° and J° be the correspond- ing open intervals with the same endpoints. Suppose that /:/->/ is continuous on I and that /=/(/). If /is differentiable on 1° and £>/(*) >0 ixEI") then f' v :J^I exists and is continuous on /. Also / "' is differentiable on J" and 112 Monotone functions Df~\y) = l (ye J") as/z->0. Df(x) provided that y = fix). Proof It follows from theorem 12.7 that /strictly increasing on /and hence/"' :/-*•/ exists. The continuity of/" 1 has already been discussed. There remains the consideration of its derivative. Let y G/°. Then x =/"'(j) S 1°. Put k =f~\y + h) -f~\y). Then /"'(;' + h) =f-'(y) + k=x + k. Thusj> + h = f(x + k) and it follows that h =f( x + k)-y =f(x + k)-f(x). Since/" 1 is continuous on J, k ■* as h ■* 0. Also,/" 1 is strictly increasing and so k & unless h =0. We may therefore appeal to theorem 8.17(ii) to obtain r l (.y + fQ-r l (y) m k | i h f{x + k)-f{_xff\x) Note that all the above results apply equally well to strictly decreasing func- tions. 12.11 Roots As an application of our observations on inverse functions, we now dis- cuss the theory of nth roots. Let n be a natural number and consider the function /defined on [0, °°) by fix) = x". Since /(0) = and f(x) -* + oo as x -> + °°, it follows from the continuity property (theorem 9.9) that the range of /is also [0, «). Observe that /'(*) = nx"- l >0 (x>0). Thus/is strictly increasing on [0, °°) and, as in theorem 12.10, it has an inverse function/" 1 . We write Monotone functions 113 This argument shows that, given any y > there exists exactly one x > (namely x =y u ") such that.y = x". Our basic assumption about nth roots made in §1.9 has therefore been justified. In exercise 10.15(1), we calculated the derivative of the nth root function. We can therefore check the validity of the formula of theorem 12.10 in this case. ^ 1/n = sr- = ^ — i — _ i y i/« y -i n{y Un f-' n y y ' If m is an arbitrary integer, the rule for differentiating a composite function (theorem 10.13) then yields D(y m ' n ) = D(y m ) iln = ^() Un y -m my m-l = ^^m/n^-1 12.12 Exercise (1) Suppose that /increases on the compact interval [a, b] . Prove that / attains a maximum at b and a minimum at a. (2) Let n G N. Prove that the function /: [0, °°) ■* R defined by f( X ) = (x + l) 1 \iln _ x l/n decreases on [0, °°). (3) Suppose that /is differentiable and increasing on an open interval /. Prove that f\x) > for each x G /. If /is strictly increasing on /, does it follow that /'(*) > for each x G /? Justify your answer. (4) Let /: R -* U be defined by f(x)= I + x + x 3 . Show that / has an inverse function/" 1 : R-*R . Calculate the value of £>/"■( y) when y=-\. T (5) A function / increases on the interval / and for each a and b in / it is true that, if X lies between f(a) and fib), then a % G / can be found such that /(?) = X (see corollary 9.10). Prove that /is continuous on /. Does the same conclusion hold if the hypothesis that /increases is abandoned? t(6) Let/: [0, 1] -*-(0, °°) be continuous on [0, l].LetM: [0, 1] ->(0,°°) be defined by M{x) = sup f(y) (0<x<l). ^y <x Prove that the function <*>(x) = lim Mix) is continuous if and only if/ increases on [0, 1]. 1 14 Mono tone functions 1 2. 1 3 Convex functions Let /be defined on an interval /. We say that /is convex on / if it is true that, for each a > and (3 > with a + (3 = 1 , 1 f(ax+py)<af(x)+0f(y) (1) whenever x G/andy E.I. If the inequality sign is reversed, we say that /is con- cave on /. It is obvious that /is concave on /if and only if —/is convex on /. The geometric interpretation is that any point on a chord drawn to the graph of a convex function lies on or above the graph. a/(.v)+i3/0-) /(O.Y+0V) /convex / concave It is sometimes useful to rewrite the definition of a convex function in the following way. A function /is convex on an interval / if and only if, for all pointsx,,x 2 and x 3 on the interval / satisfying x, <x 2 <x 3 , /(*i)-/(*i) ^ /(* 3 )-/(* 2 ) Equivalently, a function /is convex on an interval / if and only if for all points x lt x 2 andx 3 in the interval satisfying Xj<x 2 <x 3 , f(X2)-f(x 1 ) < f(x 3 )-f(x l ) (2) Xo-X, x*—x. (3) -v 3 .v, (2) slope P,P,< slope P,P 3 (3) slope P,P, < slope P,P S That (2) and (3) have the same content as (1) is easily seen by making the substitutions x, =x,x 2 = ax + (Sy and x 3 =y (see exercise 1 2.21(2)). Monotone functions 12.14 Tlieorem Suppose that /is convex on the open interval /. Then the h _>o- h >>-»o+ « both exist for each x£/. Proof Suppose that < ft, < ft 2 . Put x = x,, x 2 = x + ft, and ;c 3 = x + ft 2 in (3). Then /(x + ft,) -/(*) /(x + ft 2 )-/(x) ft 2 Hence the function F(ft) = /(x + ft)-/(x) increases in some interval (0, 6). The existence of lim FQi) h-*a+ then foUows from theorem 12.4. A similar argument establishes the existence of the left hand limit. 12.15 Example The function/: U^U defined by /(x) = \x I is convex on R . (Tlie chords lie above the graph). Note however that the two limits of (4) are not equal when x = 0. J- = 1x1 12.16 Theorem Suppose that /is convex on the open interval /. Then /is continuous on /. Proof From theorem 12.14, ( , f(x + ft )-/00 lim {/(* + ft) -/(*)} = hm 7— h->o- '•— ■"- h-*0- lim ft) =0. h->-0- Similarly, lim {/(x + ft)-/(x)} - h-*0 + ,. f(x + h)-f(x ) lim : h-»0 + lim ft| = 0. h-*0+ 116 Monotone functions 12.17 Example The function/: R-> R defined by 1 (|x|<I) /(*) = 2 (|x|>l) is convex on the compact interval [—1,1]. Observe that it is not continuous on [— 1 , 1 ) . E%%% <//////& Example 12.15 shows that a function /which is convex on an open interval / need not be differentiable on /. However, iff is differentiable on /, then the following theorem provides a useful characterisation of convexity. 12.18 Theorem Let / be an open interval and suppose that /is differentiable on /. Then /is convex on /if and only if Df increases on /. Proof (i) Suppose first that Df increases on /. Let x, <x 2 <x 3 . By the mean value theorem - /(£)> - = / (t?) X 2 ~X, X-i—X-, where x, < % <x 2 <r\<x 3 . Since Df increases, /'(?) </'(t?) and therefore inequality (2) holds. Thus /is convex on /. (ii) Suppose next that / is convex on /. Let x , < x 2 < x 3 < x 4 . By inequality /(*a)~/(*i) < /(* 3 ) -f(x 2 ) /(x 4 )-/(x 3 ) x-> — x, X,—Xi Xa Ignore the middle term in this inequality and let x 2 -> x , + and x 3 -> x 4 — . We obtain /'(x,) </'(* 4 ) and it follows that Df increases on /. If /is twice differentiable on /, an even simpler characterisation of convexity results. 12.19 Theorem Let / be an open interval and suppose that /is twice diffe- rentiable on /. Then / is convex on / if and only if f"(x)>0 for all x e I. Proof The theorem follows from theorems 12.7 and 12.18. Monotone functions 117 12.20 Examples The functions/,^ and h defined on (0, °°) by /(x) -x\ g(x) = 1/x and h(x) = x+ x~ l ate all convex on (0, °°) because their second derivatives are all non-negative on (0, °°). The function 4> defined on (0, °°) by 4>(x) = s/x is concave on (0, »). We <p"(x) = -^' 3/2 <0 (*>0). 12.21 Exercise (1 ) Let /: R -* R be defined by f(x) = (x - l)(x - 2)(x - 3). Find the ranges of values of x for which /is (a) increasing, (b) decreasing, (c) convex, (d) concave. Draw a graph. (2) Let /be convex on the interval / and let x x , x 2 and x 3 be points of / which satisfy x, <x 2 <x 3 . Prove that /(*a)-/(*i ) < /(*3)-/(*i) < fix 3 )-f(x 2 ) x 2 -x x X 3 -Xi x 3 -x 2 (3) Let /be convex on the open interval /and let J = /(/). If /is strictly increasing on /, show that/" 1 is concave on J. What happens if /is strictly decreasing on 11 *(4) Suppose that /is convex and differentiable on the open interval /. If % G /, prove that f(x)-m)>f'(%\x-%) (xe/). Interpret this result geometrically. What result is obtained if /is con- cave on/? Let 0: U -*■ R be a strictly increasing, convex, differentiable function on R and suppose that 0(£) = 0. If x, > % and = x„ — 0(*n) (a = 1,2,...) prove that x„ •* $. as n ■* so. (This method of obtaining an approxima- tion to the zero of is called the Newton-Raphson process.) + (5) Let /be differentiable, convex and bounded on R. Show that /is a constant. "I" (6) Let /be continuous on an interval /and satisfy / 'x+y , m+f(y ) 2 for each x E I and y E I. Prove that, for any x , , x 2 , interval /, , x„ in the /(— "I < ~ ifiXx)+f{X2) +■■■ +«*»)>. 118 Monotone functions [Hint: recall the induction argument of example 3.10.] Deduce that/ is convex on /. [Hint: establish inequality (1) of §12.13. Begin with the case when a and are rational.] 1 13 INTEGRATION 13.1 Area Suppose that /is continuous on the compact interval [a, b] . Does it make sense to discuss the 'area' under the graph of/? And, if so, how can we com- pute its value? If it makes sense at all to talk about the area under the graph, then presum- ably this area must be at least as big as the areas S] and S 2 which have been shaded in the diagrams below. Let S denote the set of all numbers S which can be obtained as the sums of the areas of little rectangles as in the diagrams above. Then the 'area' under the graph of /must be at least as big as every element of S. It seems reasonable to identify the 'area' under the graph of /with the smallest number larger than every element of S, i.e. with sup S. These remarks are intended to motivate the formal mathematical definition given in the next section. 119 120 Integration 13.2 The integral Suppose that /is continuous on the compact interval [a, h]. We pro- pose to give a definition of the integral f> dx of /over the interval [a, b]. We begin by defining a partition P of the interval [a, b] . This is a finite set P= [vo, y\, ■ ■ ■ ,y n ] of real numbers with the property that " = yo<yi<yi< ■ ■ -<y n -i<y n = b. Since /is continuous on [a, 6], it is bounded on [a, b].It therefore makes sense, given a partition P = {y Q ,y u . . . ,y n } of [a, b],to define m, = inf. f(x) y„«S x < y, ™ 2 = ^"t ft*) and so on . For each partition P, we can then form the sum S(P) = E^feOft-^-i). k = i The diagram below illustrates the case n = 4. The value of S(P) is the area of the shaded region. >'i >'2 j'j y* Suppose that / is bounded above on [a, b] by H. Thus /(/) <Hfor any r S fa, ft] . Then S(f) = E WfcO'fc-J'fc-i) <^ ZOfe-^-i) = -W{Oi-^o) + ... + 0' n -/„-:)} fe = i = H(y n -y ) = H(b-a). Integration 121 It follows that the set of all numbers S(P), where P is a partition of [a, b\ , is bounded above by H(b —a). Hence it has a smallest upper bound and we may define f f(x)dx = sup5(P) (1) Ja p where the supremum extends over all partitions P of [a, b] . We wish to think of J>> dx as the 'area under the graph off. But notice that any area below the x-axis has to be counted as negative for this interpretation. Note also that, if b > a, we define C f(x)dx = -f*f(x)dx. 1 3 .3 Some properties of the integral Much of the power of the idea of integration lies in its connexion with differentiation. For example, it would be very painful indeed if all integrals had to be worked out directly from the definition. Our first priority is therefore to prove the theorems which link integration with differentiation. Before we can do this, however, we need to derive a few basic results directly from the definition. 13.4 Proposition Let /be continuous on [a, b] with maximum M and minimum m. Then m (b-a)<{ f(x) dx<M(b-a). The proof of this proposition is essentially contained in §13.2. We quote two simple corollaries. 13.5 Corollary If c is a constant, then "b cdx = c(b—a). 13.6 Corollary Let /be continuous on [a, b] and satisfy |/(r)l < k for any t £ [a, b] . Then, for any £ and x in the interval [a, b] , JlfiOdt <k|x-|| (x*&. 122 Integration The next proposition is also very simple, provided that one recalls the proper- ties of sup and inf. (See exercise 2.13(2) and exercise 7.16(6i).) 13. 7 Proposition Let /be continuous on [a, b\ and let c be a constant. Then J„ {f(t) + c}dt = \ b a f(t)dt + c(b-a). The next and final result of this section is not quite so easy and so we give the proof. 13.8 Theorem Let /be continuous on [a, b] and let a<c< b. Then j a f(t) dt = £ f(f) dt + j fit) dt. Proof We use the notation of § 13.2. Suppose, in the first place, that Pi and Pi are any partitions of [a, c] and [c, b] respectively. Then the set P of points in at least one of the sets P, and P 2 is a partition of [a, b] . „D " i i i i i i ii i L p, />, It is obvious that S(P) = S(P t ) + S(f> 2 ). Moreover, ~b S(P)<j a f(t)dt because of definition (1) of §13.2. Thus, given any partition P x of [a, c] and any partition/^ of [c,b], siPi) + s(p 2 ) <[/•«) dt S(Pd<j*f(t)dt-S(P 2 ). (2) Hence, given any partition P 2 of [c, b], the right hand side of (2) is an upper bound for the set of all numbers of the form S(P{) where />, is a partition of [a, c] . Therefore sup5(P,)< f f(t)dt-S(P 2 ) where the supremum extends over all partitions P t of [a, c] . Recalling the definition of an integral, we obtain and so Integration £ fit)dt< j b a f(t)dt-S(P 2 ) s(p*)<j b a mdt-\ L a f(t)dt. A similar argument now yields J" c " fit) dt < £" fit) dt - jl fit) dt 123 and therefore •b j b a f(t)dt>f a f(t)dt + j € f(t)dt. (3) Now let P be any partition of [a, b] and let Q be the partition obtained from P by inserting the extra point c (if it does not already belong to P). It is easily seen that S{P) < 5(0. Let Pi be the partition of [a, b] consisting of those points of Q which lie in [a, c] and let P 2 be the partition of [c, b] consisting of those points of Q which lie in [c,b]. a [ II HI I 11 1 I I I I ],, We have SiP)<S(Q) =SiPd+Sm <j° a fit)dt + j* fit)dt. The right hand side of this inequality is therefore an upper bound for the set of all numbers S(P) where P is a partition of [a, b] . Hence sup S(P) < f a f(t) dt + j c f(t) dt i .e . j a fit) dt < j C a fit) dt + £ fit) dt. (4) Combining (3) and (4), we obtain the conclusion of the theorem. 124 Integration 13.9 Differentiation and integration Let /be defined on (a, b). Suppose that F is continuous on [a, b] and differentiable on (a, b) and satisfies F\x) = fix) for each x G (a, b). Then we shall say that F is a primitive (or anti-derivative) of /on [a, 6]. 13. 1 Example lff(x) = x 3 , then the function F defined by F(x) = Jx" + 3 is a primitive for/ 75.77 Theorem Let fbe a primitive for/on [a, b] and let G be defined on [a, b] . Then G is a primitive for/on [a, b] if and only if, for some constant c, G(x) = F(x) + c for each x G [a, b] . Proof Obviously F(x) + c is a primitive for /on [a, 6] . Suppose there- fore that G is a primitive for /on [a, b] . Then F-G is continuous on [a, b], differentiable on (a, b) and, for each x G (a, b), D{F(x)-G(x)} = F'(x)-G'(x) = /(*)-/(*) = 0. From theorem 1 1 .7 it follows that F — G is constant on [a, b] . Hence the result. What have primitives to do with integration? Suppose that/is continuous on [a, b] . Then we may define a function Fon fa, b] by %) - J* f®4t (a<x<b). In the next theorem, we shall show that F is a primitive of/ on [a, 6] . But first we give a rough, intuitive argument to indicate why one might suppose this to be true in the first place. Given x G (a, b), we seek to explain why one might suppose that F'(x) -f(x). The number F(x) may be interpreted geometrically as the 'area under the graph of /between a and x'. The shaded region in the diagram below should therefore have area F(x + h)—F(x). /•(.v + h)-F{x) x x + h ft Integration 125 Since /is continuous, one feels that, if h is very small, the shaded region will 'nearly' be a rectangle and hence its area will be approximately hf{x). Thus and F(x + h)-F(x) = hf(x) F(x+h)-F(x) . = fix). (5) This equation is only approximate, but the smaller h becomes the better the approximation should be. Allowing h -* 0, we might therefore expect that F'(x)=f(x). This argument is dubious on a number of counts - particularly at the stage where we divide by the small number h to obtain (5). In the next theorem we provide a precise proof of the result. 75.72 Tlieorem Suppose that /is continuous on [a, b] and that F is defined on [a, b] by F(x) m f* f{t) dt {a<x< b). Then F is a primitive of /on [a, b], i.e. F is continuous on [a, b], differenti- able on (a, b) and F\x) = f(x) for each x G (a, b). Proof Since /is continuous on [a, b] , it is bounded on [a, b] . Suppose that \f{t)\<K (a<t<b). By theorem 1 3 .8, for any x and % in [a, b] , F(x)-F® = }*/(') dt and therefore, by corollary 13.6, \F(x)-F(£)\<k\x-$\. And, from this inequality, it follows that F is continuous on [a, b] . A more subtle argument is needed to show that F is differentiable on (a, b). If x and | are in (a, b) and x ¥= %, then The last step is justified by proposition 13.7 withe =/(£). 126 Integration Let e>0 be given. If £G (a, b), then /is continuous at the point % and so, for some 5 > 0, 1/(0-/(91 <« provided that \t — £| < 5. Hence, provided that \x — %\ < 5, it follows that 1/(0 —/(I) I < e for every value of t in the compact interval with endpoints £ and*. From corollary 13.6, we conclude that F(x)-F® x-% -/(I) 1 < I* -SI 1 .elx-il = e provided that < \x — £|< 5. But this is what is meant by the assertion F(x)-FQ) *-i •/($)as*-*{. We have therefore shown that F is differentiable on (a, b) and F'(|) = /(£) for each | £ (a, 6). 75. 75 Example D c dt l+t 100 1 +x 100 ' i5.i4 Theorem Any function /which is continuous on [a, b] has a primitive on [a, b] . If G is any primitive of/, then |* /MA = G(b)-G(a) = [G(t)] b a . Proof Let Fbe defined on [a, b] by F(x) = f /(r) dt (a < a: < 6). Then J a /(/)A = F(b) = F(b)-F(a). By theorem 13.12, Fis a primitive of/ on [a, 6] and, by theorem 13.11, any other primitive G satisfies G(x) = F(x) + c (a<x<b) Integration 127 for some constant c. Hence •b J /(f) dt = {F(b) + c}- [F(fl) + c} = G(b)-G{a). 13.15 Example Since 1 D - 2(1 + ; 2 ) (1 + r 2 ) 2 ' it follows from theorem 13.14 that i; o.+ff :dt = 1 2(1 + f) 4 4 13.16 Riemann integral The integral •b /. /w* (6) was defined as the supremum of all areas like that shaded in the left hand dia- gram below. It would be equally sensible to define the integral as the infinum of all areas like that shaded in the right hand diagram below. If /is continuous on [a, b] , these two definitions yield the same result. The reason is simple. The use of the second definition would yield the result -j" -f(x)dx. But (6) and (7) are equal because F is a primitive for/ on [a, b] if and only if — F is a primitive for — /. If /is not continuous on [a, b] , the two definitions do not necessarily yield the same result. But, //they do yield the same result, we say that /is Riemann (7) 128 Integration integrable and call the common number obtained from the two definitions the Riemann integral of /on [a, b] . 13.17 Example We know that JJ X *dx = [W]h = i. To illustrate the idea discussed in § 13.16, we evaluate the integral without using the theory of differentiation. Introduce the partition P n = {0, \/n, 2/n, 3/n, . . . , 1} of [0, 1] . The shaded area of the left hand diagram is smaller than the integral and the shaded area of the right hand diagram is larger. Thus fit \ n ) n Jo tx \n] n ^{0 2 +l 2 +... + («-l) 2 }<^ x 2 dx< L{ l 2 + 2 l +...+„*} - 3 .-(n-l)n(2n-l)< f ' x 2 dx <- 3 .-«(« + l)(2#i + 1) » o ju no Considering the limit as n •* °° yields a £fe<f ll+^l ll+ ! (exercise 3.1 1(1 i)) Integration 129 13.18 Exercise (1) Prove that, given any natural number «, J>" dx = 1 B-f 1 Show that the same result holds if?? is replaced by any rational number a =£ — 1 provided that < a < b. (2) Since the integrand is a square, the following result must be wrong. Explain why. 2 f 2 ax Jo (x-l? 1 x-l = -1-1 = -2. (3) If a is a positive rational number, prove that Hffl -ir{l a + 2 a + 3 Q + ...+«"} = " - ■» « a+ 1 13.19 More properties of the integral In § 13.3 we mentioned some properties of the integral. Theorem 13.8 is particularly important but its proof which we obtained directly from the definition was long and laborious. We used the properties examined in § 13.3 to prove the theorems which connect integration with differentiation. Having proved these theorems, we can now obtain the rest of the properties of the integral very easily. 13.20 Theorem Suppose that /and g are continuous on [a, b] and that X and H are any real numbers. Then \" {x/(o + ng(t)} dt = x \ b a f{t) dt + n j* m dt. Proof Let F and G be primitives of/ and g respectively on [a, b] . By theorem 10.9(i) the function H = XF + jdG is a primitive of X/+ fig on [a, b] . Hence, by theorem 13.14, f " {X/(0 + pgffi dt = [XF(t) + M G(r)] b = HF(t)]Z + n[G(t)] b a = \f*f(t)dt + vj b a g(t)dt. 13.21 Theorem (integration by parts) Suppose that / and g are continuous on [a, b] and have primitives F and G respectively on [a, b] . Then 130 Integration \" b f(t)G(,)dt = [F(r)G(r)]2- P F(t)g(t)dt. Ja Ja Proof By theorem 10.9(h), D(FG) = fG+Fg and thus fG is a primitive of /G + Fg on [a, b\ . Hence H {f(t)G(t) + F(t)g(t)}dl = [F(f)G'(/)]* and the theorem follows. As is clear from its proof, the formula for integrating by parts is just the integral analogue of the formula for differentiating a product. The next result is the integral analogue of the formula for differentiating a composite function, i.e. a 'function of a function'. It is the rule for changing the variable in an integral and is usually remembered in the form dt dt = — . du. du 13.22 Theorem Suppose that </> has a derivative which is continuous on [a, b] and that /is continuous on an open interval which contains the image of [a, b] under 0. Then Proof Let From theorem 10.13 and theorem 13.12, d du F(0(«)) = F'(<p(u))<p'(u) = /(0( W ))0'( W ). It follows from theorem 13.14 that j a f(<p(u))<p'(u) du = [F(0( W ))]£ and this is what had to be proved. 13.23 Theorem Suppose that /and g are continuous on [a, b] and that, for any te [a, b],f(t)< g(t). Then ( b f(t)dt<fg(t)dt. J a Ja Integration 131 Proof Let H be a primitive of g — /on [a, b) . Then DH(t) - g(t) —f{t) > (a < t < b). By theorem 1 2.1,11 therefore increases on [a, b] . Thus Hib) > //(a) and hence f {git)- fit)} dt = Hih)-Hia)>0. J a 13.24 Theorem Suppose that /is continuous on [a, b]. Then f b fit)dt<[* 1/(01 dr. Proof Since - |/(/)| < fit) < 1/(01 (a<t< b), it follows that - [* |/(0I dt < J"* fit) dt < J a 6 1/(01 dt and hence the result. (Note that |/| = {/ 2 }" 2 and is therefore continuous.) 13.25 Theorem (Cauchy-Schwarz inequality) Suppose that / and g are con- tinuous on [a, b] . Then Proof This is much the same as that of example 1.11. For any x, 0<J fl " {xfit)+git)fdt = x 2 f {fit)} 2 dt + 2x\" fit)g(t) dt + I*" {git)} 2 dt JO J Q JO = Ax 2 + 2Bx + C. Thus the quadratic equation Ax 2 +2Bx + C = cannot have two (distinct) roots. Hence B 2 <AC as required. 13.26 Exercise (1) Suppose that g is continuous on [a, b] and thatg(0 >0(«<f <S). Prove that \\it)dt = if and only if git) = for each r £ [a, b] . 132 Integration (2) Suppose that /is twice differentiable on fa, b] and that/" is continu- ous on [a, b] . Prove that j b a xf"(x)dx = {bf\b)-f{b)}-{af(a)-f{a)}. *(3) Let /be positive and continuous on [1 , °°). Suppose that Prove that /(.v) > § (x - 1 ) (x > 1 ) . [Hint : The integral ft {F(t)}- 1/2 F'(t) dt is relevant.] *(4) Suppose that /and g are continuous on [a, b] and that g(t) > (a<r< b). Prove that l a mg<?)dt ~r®£ g(t)dt for some £ G [a, b] . [Hint: continuity property.] What has this to do with theorem 1 1 .6? *(5) Suppose that g is continuous on [a, b] and that /is differentiable on {a, b] and its derivative is continuous on [a, b] with f'(t) > (a<t<b). Prove that /„ f(t)g(t) dt m f(fl) j a g(t) dt + f{b) J* g(t) dt for some £ G [a, b] . What happens if f'(t) <0(a<t<b)l [Hint: inte- grate by parts.] (6) By integrating many times by parts, show that the error term in Taylor's theorem (theorem 1 1 .10) can be expressed in the form *« - &=w£ fr-o*- 1 /^) dt provided that/ (,,) is continuous on /. Show how the form of the error term given in theorem 11.10 can be obtained from that above. 13.27 Improper integrals We write down some sample definitions. Suppose that /is continuous on [0, °°). Then we define j; D °/w^ = x iimj o x /w^ provided that the limit exists. Similarly, if /is continuous on (a, b] , then we define Integration 133 T f(x)dx = lim \" f(x)dx J-i-a y-*a* Jy provided that the limit exists. Note that, if/is continuous everywhere, we define P~ f(x)dx = f° f(x)dx + P~ f{x)dx provided that the appropriate limits exist, i.e. we allow the upper and lower limits to recede to +°° and -°° respectively independently. We do not define the improper integral by lim f(x) dx. ■ — > DO J—X 13.28 Examples o>r dx „2 rx dx lim I —s = hm x->~Ji x 1 x->°> x -a. 1 -? r 1 - Jy X dx 2 On the other hand, if 0<y< 1, — = l-»+co a sy->0 + x y y and hence the improper integral j — does not exist. (ii)f 1 ^= lim f*- lim[2x" 2 ]i= Hm (2-2V>') = 2. X 1 J? ^777 v 1 f ' dx J V* ,/ — ► 1 I (iii) Note finally the following example. Observe that / ~ x dx does not exist (why not?). Hence • fjT.I * <& cannot exist even though |**xdx = U 2 -^ 2 ^0asAT^ + °°. The integrals defined in this section are called improper because they are not given by the definition of § 13.2. It is perhaps pedantic to include the little 134 Integration arrows in the notation for improper integrals and most authors would simply delete them. But for those who propose to study integration more deeply, it is probably better not to do without the little arrows since this will avoid con- fusion with the differing definitions of the powerful Lebesque theory of integra- tion. 13.29 Proposition Suppose that is a function which is continuous and non- negative on the open interval /. Suppose also that /is continuous on /and satis- fies \m\<m (xei). If the improper integral of <p over the interval /exists, then so does that of/. This is, of course, an analogue of the comparison test (theorem 6.15), for series and may be proved in much the same way. If/ happens to be non-negative on / a simpler proof may be based on theorem 12.4 concerning the convergence of monotone functions. 13.30 Example To prove the existence of the integral dx f-»oo Ji r + x 2 we simply have to observe that, for any x > 1 , dx 1 2^32- 1 +X' .-2 Since / "*~ x 2 dx exists, the result then follows from proposition 13.29. 13.31 Euler-Madaurin summation formula We discuss only a simplified version of this useful result. 13.32 Theorem Let /be continuous, positive and decreasing on [1 , °°). Then the sequence <A„> defined by An - t f(k)-C f(x)dx is decreasing and bounded below by zero. Hence it converges. Proof Since /decreases, for any k~ 1, 2, . . . , •ft + i /(* + !)< J/ f(x)dx<f(k) Thus Integration 135 a„ +1 -a„ = zm-j** 1 fix)**)- £/<*>-JT *«* = f(n + 1) - J"" + ' fix) dx </(« 4- 1) -f(n + 1) = and so <A„) decreases. Also > £/(*)- I A*) = /(«)> o fe=i fc=i and so <A n > is bounded below. From theorem 13.32 it follows that, if /is positive, continuous and decreas- ing on [1 , °°), then the series and the improper integral f A«) and C~ ftx)dx n = l Jl either both converge or else both diverge. The theorem therefore provides a criterion for the convergence of a series or an improper integral and for this reason is often referred to as the integral test. 13.33 Example Take fix) = x" 1 in theorem 13.32. We know from theorem 6.5, that the series S" =1 n~ l diverges to +°°. From theorem 13.32, we conclude that r "dx n x ■ + °° as n -* °°. In particular, the improper integral / ^°° x" 1 dx does not exist. 13.34 Exercise (1) Discuss the existence of the following improper integrals. ... r -i dx ,.., f-" x 2 dx W Jo V^7) (ll) Jo ^^* «£ V(l -x) h dx (1 -x} 3/2 (v) (1 +x 3 ) 2 *i dx «r i 1 + X + x 1 + x+x s dx o il-x) (vi) — J^o X [Hint: for (v) and (vi) recall example 13.33.] (2) Prove that, for 0< y <\, 136 Integration r(i'2)+y 2x — \, — -dx = 0. J(l/2)-y x(l —X) Discuss the existence of the improper integral J-, o ■*i 2x- ■dx. (3) Let a be a rational number with a > 1 . In theorem 6.6 it was shown that the series oo 1 converges. Base another proof of this result on theorem 13.32. 14 EXPONENTIAL AND LOGARITHM 14.1 Logarithm We define the logarithm for each x > by r* dt f * dt logx=J i - It follows immediately from the definition that log 1 = and that, for each x>0, Also, Dlogx = -. x D 2 \ogx = =. We conclude that the logarithm is strictly increasing and concave on (0, °°). Further, in view of example 13.33 and exercise 13.34 (lvi), and log x -*■ + °° as x -*■ + °° log x •* — °° as x -*■ +. (For an alternative proof of these results, see exercise 14.3(1) below.) With this information the graphs = log x can easily be drawn. The number e is defined by the equation 137 138 Exponential and logarithm log e = 1 . The value of e is approximately 2-718 (see exercise 15.6(4)). 14.2 Theorem Suppose that x > 0, y > and r is rational. Then (i) \ogxy = log* + log .y (ii) logOO = r]ogx. Proof (i) For a fixed value of y > 0, consider the function f(x) = lOgATJ-logJC. We have xy xxx for each* >0. Thus /is constant (theorem 11.7) and log xy—\ogx = c (x> 0). To obtain the value of c, put x — 1. Then c = log y -log 1 = logj\ (ii) Consider the function fix) = log(* r )-rlogJC. We have f\x) = - r .rx r -'-- X X 0. Hence /is a constant and \ogix r )-r\ogx = c. To obtain the value of c, put x = 1 . Then c = log 1 —/-log 1 = 0. Note The logarithm we are discussing here is the 'natural logarithm'. This is sometimes stressed by writing 'In' rather than 'log'. In pure mathematics, little occasion arises for the use of 'logarithms to the base 10' as found in the familiar logarithmic tables. 14.3 Exercise (1) Prove that log 2 > 0. By considering log 2" and log 2~", show that log x -* + °° as x -* + °° and that log x -*■ — °° as x -* +. (2) Prove that, for each y > 0, logy <.y — 1 . If x is a positive rational number and x > 1 , deduce that Exponential and logarithm 139 logx^y. Given a positive rational number r, prove that (i)x" r logx ->0 asx -> + <» iii)y r logy ->-0 as>> -»0 +. (These results are usually remembered by saying 'powers drown logarithms'.) [ Hint : take < s < r when proving (i).] "(3) Show that Fix) = x log x - x is a primitive for log x on (0, °°). Deduce that i log(l+*)dx = 2 log 2-1. Hence show that C W» 1 (2»)! log-' n w! - log — as n -*■ °°. e [Hint: recall example 13.17.] "(4) Show that the equation log x-ax has solutions if and only if ae < 1 . Show that elogx<x (x>0) with equality if and only if x = e. Suppose that x x > e and x n+1 = e log x„ (« = 1 , 2, . . .). Prove that x„ -*■ e as « -» °°. *(5) Prove the existence of a real number y (Euler's constant) with the pro- perty that 1 H 1 1- ... -I log n -» 7 as « ■* °°. [Hint: theorem 13.32.] Hence show that h » 2 3 4 (6) Evaluate (i) Z) {(log x) s } (JC > and s rational) (ii)i){loglogx} (X>1). Hence show that the series a i converges if r > 1 and diverges if r < 1. (See theorems 6.5 and 6.6 for comparison.) [Hint: theorem 13.32 again.] 140 14.4 Exponential and logarithm Exponential From theorem 12.10 we may deduce the existence of an inverse func- tion to the logarithm. We call this inverse function the exponential function and write y = exp* if and only if x = \ogy. Since the logarithm function has domain (0, °°) and range R, the exponential function has domain R and range (0, °°). y — exp .v Observe that the exponential function is strictly increasing on R and that exp x-t + coasx-^ + c* 1 exp x -*■ as x •* — °°. Also D exp x = 1 1 — y — exp*. D\ogy \/y Thus the exponential function is its own derivative. 14.5 ■ Exercise (1) If x andjc are real numbers and r is rational, prove that (i)exp(x+,y) = (exp x)(exp y) (ii) exp (rx) = (exp x) r . [Hint: These may be deduced from theorem 14.2. Alternatively, they may be proved directly. In the case of (i), for example, by differentiat- ing exp (x + y)/(exp x) with respect to x.] (2) Prove that, for any rational number r, (i) x~ r exp x-^ + ooasx-*-!- 00 (ii) x r exp x -» as x -* — °°. (Roughly speaking, 'exponentials drown powers'.) (3) Use L'Hopital's rule (exercise 1 1 .8(3)) to prove that Exponential and logarithm m ,. expx-1 . .... .. log(l +x) (i) lim — = 1 (ii) hm = 1 - x-*0 * x-»0 x (4) Suppose that h is positive and continuous on [0, °°) and that h {x)>H(x){x>Q\ where 141 H(x) = 1 + h{t)dt. Jo Prove that, for any x > 0, h(x)>expx. (5) Determine the range of values of x for which the function /: defined by /(*) = exp{18* 3 } is (a) convex, (b) concave. (6) If /is continuous and increasing on [0, °°), prove that [ n f{x)dx< £/«< ["* i f(x)dx. Jo k% J ' Deduce that n log n — n < log n\ < (n + 1) log (n + 1) - n and hence show that — < exp n < ; . n! n\ 14.6 Powers If a > and r is rational, then we have defined the expression a r . We propose to extend this definition to include non-rational exponents. If a > and x is real, we define a x = exp (xloga). But does this definition agree with our old definition in the case when x is rational? It follows from theorem 14.2(h) that, if r is rational, then exp {/-log a} = exp {log (a r )} = a r . Recall that e is defined by log e = 1 . Equivalently , e = exp 1 . Observe that e x = exp {x log e} = exp x which explains the familiar notation for the exponential function. 142 Exponential and logarithm 14.7 Exercise (1) If a and b are positive, prove the following. (i) a x * y = a x a y (ii) (ab) x = a x b x (iii) a" = \ (iv) (a x ) y = a xy . a (2) Prove that the derivative of a x with respect to x is (log a)a x . (3) Write 1 + -| =ex P n log | 1 + - and hence show that lim = e" (see example 4.19). [Hint: theorem 8.9 and exercise 14.5(3ii).] (4) Write « ,/n = exp l -\« % n and hence show that n l,n ■* 1 as n -* °° (see exercises 4.20(6) and 5.7(1)). (5) Discuss the existence of the following improper integrals. »£ dx J-+0 J-f-oo 1 + X 2 ;dx. [Hint: proposition 13.29.] ' (6) Obtain all solutions of the functional equations (i) f{x +y) = f(x) +f(y) (X 6 U,y GR) (ii) / (x +y)= fix) f(y) ( X € R,y S R ) (iii) /(xy) = /(at) + /(y) (x > 0, ? > 0) (iv) f(xy) = f(x)f(y) (x > 0, y > 0) for which the derivative of/ exists and is continuous wherever /is defined. For (ii) and (iv) you may assume that /never takes the value zero. 15 POWER SERIES 15.1 Interval of convergence A power series about the point % is an expression of the form in which x is a variable. 15.2 Theorem The set of values of x for which a power series converges is an interval with midpoint £. Proof We shall show that, if the power series converges when x =y, then it converges for all x satisfying \x—$\<\y—%\. The theorem then follows. I.v-£i Ir-fl ■< 5«-« »- «- *r If the power series converges when x =y, then a n (y — %)" -*• as n -> °° (theorem 6.9). Hence the sequence <a n (j' — £)"> is bounded (theorem 4.25). Thus, for some H, \a n {y-%) n \<H in = 1,2,...). Now suppose that \x — 1|< I y — f I. Then I* -« P = Hence y-ll <1. \a n (x-t-)"\ = l^^-O^.P^-^P" (« = 1,2,...). The convergence of the power series 143 144 Power series I fl„(jf-0" now follows from the comparison test (theorem 6.15) since 2„=oP" converges. We call the set of values ofx for which a power series converges its interval of convergence. If the endpoints of the interval of convergence are £ — R and | + R, we call R the radius of convergence of the power series. (If the interval of convergence is the set U of all real numbers, we say that the radius of conver- gence is infinite.) The proof of theorem 1 5.2 shows that a power series converges absolutely at all points of its interval of convergence with the possible exception of the end- points. At the endpoints the series may converge absolutely or it may converge conditionally or it may diverge. It is sometimes useful to observe that the radius of convergence R is given by - = limsup |a„| 1/n and R = lim The formulae are justified by propositions 6.17 and 6.18. In both cases, appro- priate conventions must be adopted if the right hand side of the formula happens to be or + °°. In the second case, of course, the sequence <|« n+1 \j\a„ |> may oscillate. If this happens, the limit does not exist and so the second formula becomes useless. 15.3 Example We list some power series about the point 0, together with their intervals of convergence. In each case, the radius of convergence can be calculated using the second of the formulae above. In the last three cases one is then left with the problem of whether or not the endpoints of the interval of convergence belong to the interval of convergence. Power series Interval of convergence I n=o I inx) n {0} n=0 Z* n (-1,1) Power series 145 :Ti n [-1,0 [-1,1] 15.4 Taylor series Suppose that /can be differentiated as often as we choose in some open interval / containing the point g. Then its Taylor series expansion about the point £ is f ^-j^/ (B) tt) = my fe=W*=* /»+... It should not be automatically assumed that this power series converges with sum/(x). Indeed there is no reason, in general, to suppose that it converges at all (except when x = %) and, even if it does converge, its sum need not be f(x) (see exercise 1 5 .6(6) below). A function / which is the sum of its Taylor series expansion in some open interval containing £ is said to be analytic at the point £. The most natural way of showing that a function is analytic is to prove that the error term in Taylor's theorem (theorem 11.10) tends to zero as n ■* <*>. 75.5 Example It is paTfeularly easy to write down the Taylor series expan- sion of the exponential function about the point 0. If f(x) = exp x, then f in \x) = exp x. Since exp = i , the expansion is «• x y — From examples 1 5.3 we know that this power series converges for all x. Is its sum e x l From Taylor's theorem we know that x x 2 X*- 1 »* = 1 + — + — + . . . H 1! 2! (h- 1)! «! + x - n where 77 lies between and x. Hence x x 1 + — + — 4- . . .+ 1! 2! *~] = x" n\ (« - l)!j n\ ■» as n -* °° 146 Power series (by exercise 4.20(4) or by theorem 6.9). The partial sums of the power series therefore converge to e x , and thus, for all values of x. = 1 + — + — + — +. . . 1! 2! 3! 15.6 Exercise (1) Determine the intervals of convergence of the following power series » £ e 00 I nix" n=0 OH) I (-1)" (2«>! 0v) I C-I) n : («!) n 2 (2« + 1)! (y) L ^rr.x n (vi) E & (2b)! n=l V" (2) Prove that the Taylor series expansion of log (1 + x) about the point is I (- ir 1 n=l X* X X = x + — + ... 2 3 4 Using Taylor's theorem in the form given by exercise 13.26(6) show that this power series converges to log (1 + x) for — 1 < x < 1 . Hence give another proof of the identity log2 = l-i + i-|+... (see exercise 14.3(5)). (3) Prove that the application of Taylor's theorem in the form of theorem 11.10 (with | = 0) to the function log (1 + x) yields a remainder term of the form E« = (-1)' 1+7, where r? lies between and x. For what values of x is it possible to show that E n -*■ as n -* <x> without further information about the manner in which 77 depends on x and ;i? (4) Prove that 1 1 (« + 1)! (« + 2)! Deduce that ...< 1 K + 2 (n + 1)! ' n + 1 ' Power series 147 and hence show that 2-7083 < e < 2-7184. *(5) Prove that e is irrational. [Hint: assume that e = mjn, where m and n are natural numbers and seek a contradiction using the inequality of question 4.] *(6) Let/: |R-*R be defined by fix) = ,-l/x" (**0) [0 (x = 0). Show that, for any x J 2 0, -l/x' /<">(*) = P n H e where Pis, a polynomial of degree 3m. [Hint: use induction.] Using exercise 14.5(2), deduce that, for each natural number n, ^-*0asx-»-0+. x Hence show that /can be differentiated as many times as we choose at the point and/ (n) (0) = (n - 0, 1 , 2, . . .). Write down the Taylor series expansion of/ about the point 0. For what values of jc does this converge to/(x)? 15.7 Continuity and differentiation The following proposition is useful. Its proof has been placed in the appendix since it is somewhat involved. The reader will encounter the proof again at a later stage when studying 'uniform convergence'. 15.8 Proposition Suppose that the power series f( X ) - 1 4&-£r n=0 has interval of convergence /. Then its sum is continuous on / and differentiable on / (except at the endpoints). Moreover fix) = £ m£x-Qr*. 148 Power series 15.9 Example Proposition 15.8 allows us to tackle exercise 15.6(2) in an alternative way. We know that the power series I(-l)"-'- n = i " has interval of convergence (— 1 , 1 ] (see Example 1 5.3). Hence, by proposition 1 5.8, it is differentiable on (— 1 , 1) and d K-D"- 1 n-l„n-l = £ (- I)"" 1 * = 1 -x+x 2 — x 3 + .. . 1 1 +x It follows that, for each x 6 (— 1 , I ), (-!<*<!). D log(l+*)- E (-!)"- l+x l+x = and therefore that log(l+.x)= £ (-I)"" 1 — +c (-KK1) where c is a constant (theorem 1 1.7). By substituting x = 0, we see that the con- stant c — 0. Since the power series is continuous on (— 1, 1], log 2 = lim 1ob(1+jc) = lim I (-!)»-»- x ->i- \ n=l n = I (-ir 15.10 Exercise (1) Let a be a real number. By considering D L +x) -« l «(*-». ..<a-n + l) x „ { 1=0 for \x | < l.show that Power series 149 (1 + x) a = 1 + ax a ( a ~ 1 ) ..- 2! x 2 + . . . provided that | X | < 1 (general binomial theorem). (2) Prove that the power series Z n-x 2„n n=0 converges for | x \ < 1 and determine its sum. (3) Suppose that the power series f( X ) = | a n x n n=0 has interval of convergence /. If y E I but is not an endpoint of/, prove that \jMdx= 1 -Shy*. Jo n=0 n 4- 1 (4) Prove that i log ( 1 -x) dx. n=0 n (5) Suppose that the two power series n=0 «=0 converge in some open interval /containing \. Prove that their sums are equal for each x G / if and only if a n m b n (n m 0, 1,2, . . .). (6) Suppose that the power series y = fix) = t On** n=o s — - converges for all real x and satisfies the differential equation dy dx = y. l Show that a n+l = — — a n (n = 0, 1 , 2, . . .) and deduce that n + 1 f{x) - a e*. 16 TRIGONOMETRIC FUNCTIONS 16.1 Introduction We based our definitions of the exponential and logarithm functions on the formula f * dt We could equally well have begun with the differential equation %-y dx and defined the exponential function as the sum of a power series as indicated in exercise 15.10(6). In a similar way, the definitions of the trigonometric functions can be based on the formula arctan x -r Jo dt ]+t 2 ' We prefer, however, to base them on the differential equation \ + y = 0. dx* We already have some intuitive ideas about the sine and cosine functions from elementary geometry and trigonometry. It is therefore sensible to begin by indi- cating why the sine and cosine functions, as conceived of in trigonometry, should be expected to satisfy this differential equation. In the diagram, the angle x is measured in radians and.y = sin x and z = cos x. By the sort of argument considered adequate in elementary trigonometry, we obtain dy , dz v = sin x = dx = —z. 150 z = cosx = — : ; dx y = sinx = ich it follows that d 2 y _ dz dx 2 ~ dx ~ y '' d 2 z dy dx 2 " dx Trigonometric functions 151 16.2 Sine and cosine Suppose that the power series /(*) = t **" n=0 converges for all real values of x and that /"(*)+/(*) = o. By proposition 15.8, f(x) = a + a t x+ a 2 x 2 +...+ a n x n +... fix) = «!+ 2a 2 x+ 3a 3 x 2 +...+ in + l)a nti x n + . . . fix) = 2a 2 + 3 . 2a 3 x + 4 . 3a A x 2 + . . . + (n + 2)(n + l)a n + 2 x" + ... Thus, to satisfy the differential equation, a Q and a, may be chosen in any way, but then the remaining coefficients must be chosen to satisfy in + 2)in + l)a n+2 + a n = (« = 0,1,2,...). This recurrence relation is readily solved and we obtain „2n ,,2n+l '*I<- I >*e* +8 'S ei) V+»>' n=0 We want sin = and D sin = cos = 1 . We therefore define sin* ■■ £ (-1)" Similarly n=0 x 2n+l =;t _^! + (2n + l)! * 3! ^2" ««-£<-»" m 1 - ?! 152 Trigonometric functions Both these power series converge for all values of x by comparison with the power series for e x . Note that we take these formulae as definitions. The definitions of elementary trigonometry concerning the 'opposite', the 'adjacent' and the 'hypotenuse' are not precise enough for our purposes in this book. The remarks of §16.1 are simply intended to indicate why we chose to define sine and cosine as we did. 16.3 Exercise (1) Prove the following. (i) cosO = 1 (ii) sinO = (iii) cos (— x) = cos x (iv) sin (— x) = — sin x. Show that, for all values of X, (v) D cos x = — sin x (vi) D sin x = cos x. (2) Let y be any real number and define g(x) = sin (x + y) — sin x cos y — cos x sin y h(x) = cos(x +y)~ cosx cosj' + sin* sin.y. Differentiate {g(x)} 2 + {h(x)} 2 with respect to x and hence show that, for all values of* andy, sin (x+y) = sin x cos y + cos x sin y cos (x + y) = cos* cosy — sin x sin y. (3) Use the previous question to prove the following. (i) cos 2 x + sin 2 * = 1 (ii) | cos x | < 1 and I sin x I < 1 (iii) sin 2x = 2 sin x cos x (iv) cos 2x = cos 2 x— sin 2 x. (4) Use L'Hopital's rule to prove that „, sin x (i) »1 asx^-0 ... 1 — cos x 1 00 — ^ >-asx 0. trt (5) Use the mean value theorem to show that, for all values of x sinx <1. Hence show that the series 2™ =1 sin (l/« 2 ) converges. Discuss the con- vergence of S" =I sin (l/«). Trigonometric functions 153 *(6) Use exercise 13.26(5) to prove that c" sin t -dt m provided that n>m> 0. Explain why it is possible to deduce the existence of the limit "sin t (-"sin lim — „->=oJl t dt. 16.4 Periodicity Since cos = 1 and cos (— x) = cos x and the cosine function is con- tinuous, there exists a % > such that cos x > for x S (— £, £). But the cosine function is not positive all the time. If this were the case, then it would follow from the formula D 2 cos x = — cos x that the cosine function was concave. But a bounded, differentiable function cannot be concave unless it is constant (exercise 12.21(5)). It follows that there exists a smallest positive number % for which cos % = 0. We define the real number n by \n = %. By definition, cos £tt = cos (— \n) = and cos x > for - \v <x < \-n. We show that sin \it = 1 . It follows from the formula cos 2 x + sin 2 x = 1 that s i n 2 Jtt = 1 . But D sin x = cos x and so the sine function increases on [— {-n, j7rj . Since sin = it follows that sin jtt > 0. We now appeal to the formulae of exercise 16.3(2). We have sin (x + j7r) = sin x cos \ti + cos x sin jt7r = cos x cos (x + jit) = cos x cos \tt — sin x sin \tt = — sin x from which it follows, in turn, that sin (x 4- 2n) = sin x cos (x + 2n) = cosx. Because of these last formulae we say that the sine and cosine functions are periodic with period 2n, 154 Trigonometric functions 16.5 Exercise (1) Prove thai the cosine function decreases on [0, rr] and increases on [ n, 2n] . Prove that it is concave on [— \-n, \tt} and convex (2) We define the tangent function by on [2^. jt] tan* = sm x COS X (x^{n + \)rt:n = 0,±l,+2, . . .). Prove that tan (x + 7r) = tan x provided that x =£ (« + £) i\ where n = 0, ±1, ±2, . . . Show that the tangent function is strictly increasing on (— \ir, \tt) and that tan x •* + °° as x -> \tt — and tan x -* — °° as (3) Show that the sine function is strictly increasing and continuous on [ — ?"■> 5 7r ] ar| d tnat { ' ie image of this interval under the sine function is [— 1, 1]. If we ignore the fact that the sine function is defined out- side the interval [— \-n, |ir], the function we obtain therefore admits an inverse function with domain [—1,1] and range [— \ti, \-n\ . We call this function the arcsine function. Draw a graph of this function and calculate D arcsin x for — 1 <x < 1 . (Some authors use the notation sin" x, but we prefer not to, since the sine function has no inverse. The arcsine function is the inverse of the restriction of the sine function to [— \-n, ^n]). Discuss the arccosine function obtained as the inverse of the restric- tion of the cosine function to [0, rr\ . Calculate D arccos x and explain how the arcsine and arccosine functions are related. (4) Show that the arctangent function, obtained as the inverse of the' restriction of the tangent function to (— \tr, \-n), has domain U and range (— j7T, \tj). Draw a graph and calculate D arctan x. Hence show that dx Jo \+x 2 2' + (5) Suppose that /has a continuous derivative on R and that /M+/O0 =/ x + y \—xy for each x andy such that xy <\. Prove that, for some constant C, f(x) - C arctan x. By writing* =jc in the formula and considering what happens when x -*• 1 — , prove that arctan 1 = \n. Show that, for - 1 <x<l. X X X arctan x = x 1 K.. 3 5 7 Trigonometric functions 155 and hence obtain the identity J„ = !_«+£->+... t (6) Let/: fix) = *U be defined by sin - {x # 0) x (x = 0). Let#: R -> R and/i: R->R be defined by g{x) = xf(x) and h(x) = x 2 f(x). Draw graphs and prove the following. (i)/is not continuous at the point 0. Qi)glS continuous at the point but not differentiable there, (iii) h is differentiable at and h'(Q) = 0. 17 THE GAMMA FUNCTION 17.1 Stirling's formula If ( a n ) and < b„ > are sequences of real numbers, the notation a n ~ b n means that *■ 1 as n -+ ca . b„ 17.2 Proposition {Stirling's formula) nl~y/(2ir)n n n i/2 e- n . Proof Consider the sequence < d n > defined by d n = log «!—(« + j) log n+n. We seek to show that (,d n > decreases and so we examine the sign of d n -d n+l = -log(«+ 1) - (« + |) log « + (n + |) log (« + 1) — 1 4 4-4^-' 2n + l |l+(2n+l) _1 But, for |jc|<1, 1 m = ^io g i +* 1 -x 2 -1 2*11 2 3 jc 2 x A x 6 3 5 7 2 3 -1 0) In particular, f(x)>0 for U|< 1 and thus d„ ~d n+1 >0 (n = 1, 2, . . .). Therefore <c/„> decreases. 756 The gamma function From (1) it also follows that, for | x \ < 1, f(x)<j{\+x 2 +x*+...} 157 Thus 3(1 -x 2 ) 2\" rf n -"<*„+,<; 1 3{(2n + l) 2 -l} 12« 12(n+l) It follows that the sequence (,d n —(12b)" 1 } is increasing. In particular, d n -{\2n)- i >d l --h (n = 1,2,...) and hence the sequence < d n > is bounded below. Since < d n > decreases, it follows that {d n ) converges. Suppose that d n -* d as n -*■ °°. Since the exponential function is continuous at every point, exp d n •* exp d as n •*. •*>, Let C = exp d then n\ n"n U2 e- n C as n -*■ °°. It remains to show that C = V(2t)- We leave this as an exercise (exercise 17.4(1) and (2)). 1 7 .3 * The gamma function The gamma function is defined on (0, °°) by the formula r(x) = J~*V -l <r*«a («>o). To justify the existence of the improper integral, we appeal to proposition 13.29. In view of the inequality the existence of the integral £ fl t x ''e-'dt follows from that of £ f x_1 rfr, pro- vided that x > 0. Since r x+1 e"' ■* as t -> + °°, we have, for some i/ >0, f*-*«-»<JSft** (r>l) and hence the existence of r*°° t x ~ l e~'dt follows from that of/,"*" f " 2 dr, /7.4 + Exercise t(l) Let / n = /f" 3 sin" xdx (n = 0, 1 , 2, . . .). Prove that < /„ > is a decreasing sequence of positive numbers which satisfies nl„ = (n — l)/ n -2 (n = 2, 3, . . .). 158 Tlie gamma function Deduce thai hn 1 as n -> °°. ^(2) With the notation of the previous question, establish the identities (2n)\ n hn (2 n n\f 2 (2"«!) 2 '2n*l (2n+ I)! (" =0,1,...) (« = 0,1,...). Hence show that the constant C obtained in §1 7.2 satisfies C = \/(2n). T (3) Let d n = log n\ - (n + j) log n+n. Show that the sequence {d n — (12«) -l > increases but that the sequence (d n — (12n + 1) _1 > decreases. Deduce that g l/(12n-H)^ n\ \/(27r)n"n V2 e- n <P vl2n Hence estimate the error on approximating to 100! by Stirling's formula. Is this error large or small compared with the value of 100!? + (4) Show that r(x + l) = xr(x)(x>0). Deduce that, fom = 1,2,3, ... , r(n + l) = b! ' (5) Prove that the gamma function is continuous on (0, °°). [Hint: If 0<a<a<.K<_)><£<|3, prove that, for some constant H which does not depend on x aty, \r(x) - T(y) \<ff\x —y | {r(a) + 1X0)}.] ■ (6) Prove that the logarithm of the gamma function is convex on (0, °°). (Hint: exercise 12.21(6) and theorem 13.25.] 1 7.5" 1 Properties of the gamma function The gamma function provides a generalisation of the factorial function (see exercise 17.4(4)). Is it the only such generalisation? 1 7.6* Theorem Let /be positive and continuous on (0, °°) and let its logarithm be convex on (0, °°). If/ satisfies the functional equation /(x+1) = */(*) (x>0) and/(l)= l.then /(*) = T(x) (x>0). Proof The proof consists of showing that, under the hypotheses of the theorem, for each x > 0, The gamma function fix) = lim n x n\ „"'- x(x +])... (x+nY 159 (2) It follows from exercise 17.4(4, 5 and 6) that the gamma function satisfies the hypotheses of the theorem. Since a sequence can have at most one limit, we can therefore conclude from (2) that/(x) = T(x) (x > 0). The proof of (2) uses the convexity of log / Suppose that s < f < s + 1 . Then we may write t = as + (3 (x + 1) where a>O,0>Oanda + |3=l. Now t = (a + 0)s + (3 = s + and so = t - s. From the convexity of log/, it follows that log/(r) < a log/(x) + log/(s + 1) fit) < imrm + o}" = s^m = s<-*m. (3) Since s < t < s + 1 , we also have / — I < s < f . Making appropriate substitu- tions in (3), we obtain f(s) < (t - ir**V(* ~ 1) - fr~ l) s "'/(0- w Combining (3) and (4) yields the inequality (t-\y- s f( S )<f(t)<s'- s f(s). Now suppose that <x < 1 and that n is a natural number. We may take s = n + 1 and t = x + n + 1 . Then (x + n) x f(n + 1)< f(x +n + 1)<(» + D x /(« + 1 )• (5) From this inequality it follows that (x + n) x n ! < (x + n)(x + n - 1) . . . xf{x) <(n + l) x n\ ,^ >+.x»+ ; .-n...v M < (, + i)', This completes the proof of the formula (2) in the case when < x < 1 . The general case is easily deduced with the help of the functional equation f(x+l)=xf(x). 17.7^ Exercise + (1) lfx>0, prove that r« = r log- I t dt. 1 60 The gamma function + (2) Prove that T(x) ~ yj(2n)x x x' in e' x . This is Stirling's formula for the gamma function. [Hint: see theorem 17.6, inequality (5).] *(3) Use L'Hopital's rule to show that lim z-»0 log(l +z)-z With the help of Stirling's formula for the gamma function, deduce that, for all x > 0, (V«)/u(« +*>/«)• 1 V(2rr) e ' as u where 1 AM - j^-V. * (4) The beta function is the function of two variables defined for x > and y > by the formula B(x,y) =P' ^-'(l-ty-'dt. J-*o Check that the improper integral exists provided that x > and y > 0. Prove that, for a given fixed value of y > 0, B(x, y) is a positive, continuous function of x on (0, °°) whose logarithm is convex on (0,-). '(5) With the notation of the previous question, prove that, for a given fixed value of y > 0, the function /: (0, °°) -*R defined by f(& = r( * ( + y f B(x,y) (x>0) satisfies the conditions of theorem 17.6. Deduce that, for x > and ^>0, B(x,y) = r«ro>) r(x+y) T (6) Use the previous question to evaluate T(j). Hence show that P~ e~ x2 ' 2 dx = V2ir. 18 APPENDIX 18.1 Introduction In the preceding chapters a number of results were stated without proof. These results were referred to as 'propositions'. The proofs were omitted from the main body of the text to avoid confusing the issue with too much detail. Instead we give the proofs here. 1 8.2 Bounded sets Proposition 2.3 A set S of real numbers is bounded if and only if there exists a real number K such that | x \ < K for any xES. Proof By exercise 1.20(1), \x | <^if and only if -K<x<K. If | x | < K for any x e S, it follows that - K is a lower bound for 5 and K is an upper bound for S. Thus 5 is bounded. Suppose, on the other hand, that S is bounded. Let H be an upper bound and h a lower bound. Put K = max {| H I, I h !}. Then -K<h<x<H<K (xGS) and hence | x \ < K for any x G 5. 1 8 .3 Combination theorem Proposition 4.8 Let x n -» / as n ■* °° and y n ■* m as n -* °°. Let X and m be any real numbers. Then (i) Xx n + p.y n -*\l + /t»i asn-*<* ('0 Vn -*lm as n-*°° x I (iii) — -> — as n -* °° (provided that m i= 0). y n m Proofs o/0'O and (iii) (ii) Since <x„) converges, it is bounded (theorem 4.25). Suppose that !*„!<*(» = 1,2, 3,...). Consider 161 162 Appendix \x n y n -lm\ = \x n y n -x n m+x n m-lm\ < I x n I • I y n ~~ m I + I m I . I x n — 1 1 (triangle inequality) < K.\y n -m\ + \m\.\x n -l\ = z n - But x„ -► / as « -►o and so | x„ — l\-*Q as n -►<*> (exercise 4.29(H)). Similarly | y n — m | -* as n -* °°. From proposition 4.8(f), z„ -*0 as/i -»•<». Hie result then follows from corollary 4.1 1. (iii) Since y„-*-m as« -*■**, \y„ | -> \m | as« -*°° (exercise 4.29 (lii)). Since m # 0, | m | > 0. It follows from exercise 4.29(2) that we can find an N such that, for any n >N, \y n \>h\m\. For n > N, consider X n_J_ y n m 'nx n -yj my„ < m \mx n -y n l\. By proposition 4.8 (i), mx n - y n l -*■ ml — ml = as n -* °°. The result therefore follows from corollary 4.1 1 . 18.4 Subsequences Proposition 5.13 Let (x n > be a bounded sequence and let L be the set of all real numbers which are the limit of some subsequence of (x n ). Then L has a maximum and a minimum. Proof The set L is non-empty (Bolzano-Weierstrass theorem) and bounded (theorem 4.23). Hence L has a smallest upper bound/. We seek to show that IGL, i.e. there exists a subsequence (x n > such that x n -*lasr-*°°. Let e > 0. Then je > 0. Since 7 is the smallest upper bound of L, 1 — £e is not an upper bound of L. Hence there exists znlEL such that -I, l>l>l-\e. and therefore |f-7|<ie. Because / G L, we can find a subsequence < x m > which satisfies x m -*■ I as r - Hence there exists an R such that, for any r >R, \x m — u<y. (i) (2) Appendix 163 Taking (1) and (2) together, we obtain, for any r >R, \x mr -i\ = \x mr -i+i-h < l* mr -/| + |/-7l < ke + he i.e. for any r >R, \x mr -l\<e. This does not conclude the proof since the subsequence <x mj _) will change as e changes. All we have shown is that, given any e > 0, there exists an infinite collection of terms x n of the sequence < x n > which satisfy |x„-?|<e. ( 3 > With this information, we construct inductively a subsequence (x„ r ) which satisfies x n -* 1 as r •* °°. Take e = 1 in (3). Then there exists an «, such that | *„_ - / 1< 1 . Take e = \ in (3). Then there exists an n 2 > n , such that | x„ 2 - 1 1 < k- In this wa Y we construct a subsequence <jc„ > which satisfies l*n r -'l<7 and hence x„ r -*■ 7 as r -*■ °° (sandwich theorem). A similar argument shows that the infimum / of L also belongs to L. Proposition 5.17 Any convergent sequence is a Cauchy sequence. Proof Let e > be given. Then £e > 0. If x n ■* I as n -* °°, then we can find an A' such that, for any n>N, \*n-l\<te. Equally, for any m > A^, \x m -l\<i L e. Thus, if m > N and n > N, \x n -x m \ = \x n -l + l-x m \ < \x n -l\ + \x m -l\<ke + he = e and thus ix n ) is a Cauchy sequence. Proposition 5.18 Any Cauchy sequence is bounded. Proof Let <x„> be a Cauchy sequence. It is true that, for any e > 0, we can find an N such that, for any n > N and any m > N, | x n - x m | < e. In parti- cular, this is true when e = 1 , i.e. there exists an N t such that, for any n>Ni 164 Appendix and any m>Ni. \x n -x m \<\. Take m = N t + 1 . Then, by theorem 1 . 1 8, for any n>N lt \x n \-\x Ni + l \<\x n -x Nitl |< 1 l*nl<l*JVll + !■ Now take A' = max {|*i | 7 |x 2 1, . . . , \x Ni |, |*n,-mI + 1} and it follows that | *„ | < A" (w = 1 , 2, . . .). 1 8.5 Tests for convergence of series Proposition 6.17 {ratio test) Let S~ = , a n be a series which satisfies lim <n+l = /. If /> 1, the series diverges and, if / < 1, the series converges. Proof The text indicates the proof for the case / < 1 . If / > 1 , we may take e > so small that / — e > 1 . Then, for a sufficiently large value of /V, Kv-h «n-l "n-2 "N+2 >Q-e) n-N*l a N+i I -*• + °° as M Hence a n j*Q as n -*■ °° and so 2~=i a n diverges. Proposition 6.18 (nth root test) Let 2„ =! a„ be a series which satisfies lim sup |a n | 1/n = /. If / > 1 , the series diverges and, if / < 1 , the series converges. Proof Let x n = \a n |"". If / < 1, we may choose e > so small that ' + e < 1 . From exercise 5.15(4) it follows that we can find an iV such that, for any n > N, i.e. x n < I + e \a n \ <(/ + £)". Since 2™ =1 (/ + e)" converges, the convergence of E~ =1 a n follows from the comparison test. Appendix 165 If I > 1 , we may choose e > so small that / - e > 1 . Let < x„ r > be a sub- sequence such that *„ r -> I as r -*■ °° (proposition 5.13). We can find an R such that, for any r>R, x„>Q-e) \a nr \>(l-e) n r-> + °°asr^°°. Hence a„^0asn-» M (theorem 5.2) and thus S*-i a n diverges. 1 8.6 Limits of functions Proposition 8.4 Let /be defined on an interval (a, b) except possibly at a point g e (a, ft). Then/(*) ■* I as * ■+ g if and only if /(x) -Was* ■* | -and /(*)->/ as x->£+. iVoo/ (i) Suppose that /(*) -» / as * -*• |. Let e > be given. Then we can find a 5 > such that |/(*)-/|<e provided that 0<|*-£l<5.But£-5<*<£ implies that < I * - % |< 5 (see §8.3). Hence | - 6 < * < g implies that | /(*) - / 1< e. Thus /(*) ■* I as Similarly, % < * < g + 5 implies that < I * - g 1< 8. Hence ?<*<? + 5 implies that | /(*) -l\<e. Thus /(*) -* / as * -* g +. (ii) Suppose that /(*) -* I as * -*■ % - and /(*) ■* I as * ■+ g +. Let e > be given. We can find a 5j > such that | /(*) - / ] < e provided that g-5 1 <*<|. Also we can find a 6 2 > such that | /(*) - / 1 < e provided that £<*<g + 8 2 . Let 5 = min {S, , 5 2 } . Then | /(*) -l\<e provided that £-6<*<£andg<*<g-6 i.e. i.e. ^-5<*<? + 5and*^? 0<|*-£l<5. Thus /(*)-* /as*-* |. Proposition 8.12 Let /and g be defined on an interval except possibly at £ G (a, ft). Suppose that f(x)-+l as * ■* g and g(*) ■* m as * -» g and suppose that X and n are any real numbers. Then (i) X/(*) + pg(x) -> X/ + (WW as * ■* £ (ii) /(*)£(*) ^/« as* ^£ 166 Appendix (iii) f(x)/g(x) -* l/m as x -*■ £ (provided w =£ 0). Proof (i) Let <x n > be any sequence of points of (a, b) such that x„ =£ £ (n = 1 , 2, . . .) and x„ -> £ as n -> <». By theorem 8.9, /(x„) -* / as n -*■ °° and g(x n ) -* m as « ■+ °°. By proposition 4.8(i) */(*n) + Atf(*n) -* X/ + /im as « -> °°. Appealing to theorem 8.9 again we obtain X/(x) + Mf (■*) ~* M + A"" as* •* f. Items (ii) and (iii) are proved in exactly the same way. Proposition 8.14 Let /, g and h be defined on (a, b) except possibly at £ G (a, b). Suppose that g(x) -*■ I as x -*• £, h (x) -> / as x •+ £ and that g(x)<f(x)<h(x) except possibly when X —%. Then fix) -*■ I as x -* §. Proof Let < x„ > be a sequence of points of (a, £) such that x„ =£ £ (n = 1,2, . . .) and x„ ->• £ as n ->• °°. By theorem 8.9,#(x„)->/ as «-*» and /i(*n) -> Z as n -* oo. Since g-(x„) </(x„) < /z(x„) it follows from theorem 4.10 that/(#„) -> / as n -*■ °°. Hence /(x) ■* / as x ■* £ by theorem 8.9. 18.7 Continuity Proposition 9.3 Let /be defined on an interval. Then /is continuous on / if and only if, given any x G / and any e > 0, we can find a" 5 > such that l/(*)-/O0Ke provided that y& /and \x— y | <5. Proof If *€/ but is not an endpoint of/ then the condition y G / and I x —y | < 5 is just the same as | x — y |< 5 provided that 8 is sufficiently small. The criterion given in the proposition therefore reduces to the assertion that f(y) ""*■/(*) as y •* x, i.e. /is continuous at x. Suppose that x € / and is a left hand endpoint of /. The the condition y G I and | x — y | < 5 reduces tox<y<x + 8, provided that 5 is sufficiently small. The criterion of the proposition therefore reduces to/(^) ~*f(x) asy -*x +, i.e. /is continuous on the right at x. Similarly if x G/and is a right hand end- point. Propositions 9.4, 9.5 and 9.6 are entirely trivial consequences of the results indicated in the text. Appendix 5 1 8.8 Integration Proposition 13.4 is proved in §13.2. 767 Then have Proposition 13. 7 Let /be continuous on [a, b] and let c be constant. \\ {/(<) + c} dt = £ f{t) dt +c(h- a). Proof hztP= {yo,y\,y2, ■ ■ • >y«} denote a partition of [a,b]. We m\ U+c) _ inf {f(x) + c}=c+ inf {/(*)} = e + m\ n . y <x<y, j> <x<y, Similarly mj/ +c) = c + m^ (k - 1, 2, . . . >a). It follows that s if+c) (p) m | mg* e \y k -y k _ t ) fe=i n n = E wi'^fc-^fc-J + e Z (n-yfe-i) fe=l fe=l = S in (P) + c(b-a). It follows that \ {f(t) + c} dt = sup {S if * c \P)} = sup {S«\P) + cib -a)} Ja p p = sup {S<- n iP)}+ cib -a) = l" f(t)dt + c(b-a). p Ja Proposition 13.29 Suppose that is a function which is continuous and non-negative on the open interval /. Suppose also that /is continuous on / and satisfies |/(*)|<0(X) (xG/). If the improper integral of (p over the interval /exists, then so does that of/. Proof We consider only the case when / = (0, °°) and the integral / = P°° <t>ix)dx exists. Let «n = f" /(*)<**; bn=r <Kx)dx (« = 1,2,...). Jn-l Jn-1 Then the series 2~=i b„ is a convergent series of non-negative terms (with sum /). Also 168 Appendix \a n \<b n (H = 1,2,...) and hence 2" =1 a n converges by the comparison test, i.e. ■ N m = lim f f(x)dx exists. Given any X > 0, we may take N to be the smallest natural number satis- fying N>X. Then <&;v->-0as A 7 -*- 00 . It follows that -x = »? lim f f(x)dx = as required. 1 8.9 Continuity and differentiation of power series Proposition 15.8 Suppose that the power series fix) = £ «„(*-*)" n=0 has interval of convergence /. Then its sum is continuous on / and differentiable on / (except at the endpoints). Moreover f(x)= X mdx-HT' 1 - n=l Proof Let the radius of convergence of the power series be R. Suppose that X is any point of / other than an endpoint. We can then find another point x of / so that x lies between £ and x and thus [X-f|<|ar -||<fi. -/?— ■ x- -> -f— 1- £ x v„ From the formula of §15.1 — = lim sup |a n l" n = lim sup K n~>°° n—*-°° n(n-\) l/n Appendix we may deduce the convergence of the series «(«-!) 169 I : a„*o" 2 where Xo = *o — £• Put 5 = | x — x \. For values of y satisfying < | x — y | <6, consider ^— x ^i n =i [ r— X where X = x — ? and F = y — if. We seek to show that A -* as Y-*-X. From exercise 3.11(2), n v"" - nX n ~ l = (Y"- 1 +Y"~ 2 X+ ... + YX"' 2 + X"' 1 ) - nl" = (Y"' 1 -X n ~ l ) + x(r n_2 - x"- 2 ) + ...+ X"-\Y- X). But from each of these terms we can extract a factor of (Y — X). The right hand side then reduces to the product of {Y — X) with the sum of \n(n — 1) terms of the form X r Y s where r + s = n — 2 . The number \n (n — 1 ) arises from the use of the formula 1 + 2 + 3 + .. . + (n - 1) = %n(n - 1) (example 3.9). Since \X\ = |x-5l<|x -5l - I *,, I and m - ly-|KI*-tl + » » 1-Xbl we have Hence y n -z" r-AT -nJC n-I <-«(n-i}ijr i nH »|y-jfK lAKir-jriJ^i^Ti -+0as K-i-AT. It follows that /is differentiable at each x in / which is not an endpoint and /'(.*) = £ M#**. n-1 There remains the question of left or right hand continuity at the endpoints of the interval of convergence /, if these happen to belong to /. A somewhat more subtle argument is required to deal with this question which is the subject matter of Abel's theorem quoted below. Abel's theorem If the series Sfe =0 a k converges, then 1 70 Appendix lim £ a k x k \ = £ a k . *-+l- { fe=o ) k -° Proof Let e > be given. Since the series 2 a k converges, its sequence of partial sums is a Cauchy sequence. We can therefore find an N such that, whenever k>m>N, k Z a, <r- (4) Abel's lemma (which is easily proved by induction) asserts that + v n . J u h . k-m k=m \\l=m We apply Abel's lemma with u k = a k and v k = x . Then k-m fe=m k=m It follows that, for n > m > N and < x < 1 , la k x k k=m <0-*)"f \ex k + \ex n k=m -> a 3 (1 — x) 3 3 We may conclude that I a k x' k=N < r- (5) Next observe that, for < x < 1 , £ a k x k - V a k fe=0 k=0 N-i ! j < I \a n \(\-x") + -e + -e. fe=0 This follows from (4) and (5). But the finite sum on the right hand side tends to zero as x -* 1 — . We can therefore find a 5 > such that, for any x satisfying \-8<x<\, N t\a n \(l-x")<\e. k=o 3 Given any e > 0, we have therefore found a 6 > such that, for any x satis- fying 1 - 5 < x < 1 , T Appendix 171 _ a k x k - £ tffe fe=0 ft=o This concludes the proof. <e. 18.10 Stirling's formula The proof of proposition 17.2 is completed in exercises 17.4(1 and 2). SOLUTIONS TO EXERCISES • Exercise 1.8 1 There are three possibilities: x = 0, x > 0, x < 0. In the first case x 2 = 0. In the second case rule III yields that a; 2 = x.x >0.x = 0. Rewrite the third possibility iii the form >x and apply rule IV. Then = 0.x<x.x = x 2 . In each casex 2 >0, i.e.x 2 >0 or x 2 = 0. (i) We are given that < a < 1 . Since a > we may apply rule III. Then = 0. a<a.a<l.a = a. (ii) We are given that b > 1 . Since b > we may apply rule III. Then b 2 = b.b>l.b = b. 2 Since bB >0, a. /4 ff.fi = -.bB<-.bB = Ab (rule III), o fi We deduce that a{b + B) ■ a* + flfi < aft + v46 = (a + 4)4 (rule II). Since (6 + B)' 1 > and 6" 1 > (example 1.5), it follows from rule III that a a + A b b + B' The second half of the inequality to be demonstrated is obtained simi- larly. 3 Suppose that a > b and c> d. Then, by rule U,a+c>b + c and b + c>b+d. Hence a + c>b + d (rule I). If b > and d > 0, we first observe that c > d > 0. Then, by rule III, ac > be and 6c > W, Hence ac > M (rule I). 4 Substitute the values a = 5,b = 4, c = 3,d = 1 in inequalities (i) and (ii). We obtain the false assertions 7 72 Solutions to exercises (1.8, 1.12) 173 2 = 5-3>4-l = 3 5 4 i.e. 5 > 12. Note that the inequalities are false in spite of the fact that b >0 and d > 0. On the other hand inequality (iii) is true when b > and d > (exercise 1 .8(3)). However, if we take a = c = — 1 and b = d = — 2 in inequality (ii), we obtain the false assertion 1 = (-l)(-l)>(-2)(-2) = 4. 5 We have, for any e>0,b<a + e and a — e < b. Since b < a + e for any e > 0, it follows that b < a (example 1 .7). Since a < b + e for any e > 0, it follows that a < b. Hence a = b. 6 Take x = (a + b)/2. • Exercise 1.12 1 If n is an even natural number it may be expressed in the form n = 2k. But then x" = x 2h - = (x R ) 2 >0 for all values of* (exercise 1.8(1)). Hence, if y < 0, x" = y has no solutions. The equation x" = has only the solution x = since x ¥= implies that x" =t 0. Uy > 0, our assumption about the existence of nth roots assures of the existence of a unique x>0 such that x" = y. But, since n is even, x" = y if and only if (— x)" =y. Hence the equa- tion has exactly two solutions, one positive and one negative. Next suppose that n is odd. If y = there is no difficulty in showing that*" =y has exactly one solution. Ifiy >0, there is exactly one positive solution and this is the only solution because x < implies .*" < when n is odd. If y < 0, we use the fact that z" = — y has one and only one solution and hence the same is true of (— xf = —y, i.e. x"=y. / 74 Solutions to exercises ( 1. 12 J 2 (i) 8 2 ' 3 = {(2 3 ) 2 }" 3 = {4 3 }" 3 = 4 1/3 I . \ 1/3 (ii)27 -4/3 = 1 27' 1 P 3 ) 4 J _1_ 81 (iii) 32 6/s = {(2 5 ) 6 } 1/5 = 2 6 = 64. 3 We take for granted the truth of (i), (ii) and (iii) in the case when r and s are integers. To prove the results when r and s are any rational num- bers, it is helpful to have available the following preliminary results in which m and n denote natural numbers: (a)O m ) ,/n = (y Vn ) m (b)0 1/m )"" = y umn (c)y Un z Vn = (yz) Vn To prove (a), observe that {(}> "")"*}" = (y" n ) mn = {(y 1/ ")"} m = y m and hence (y v ") m is the unique positive solution ofx" = y m , i.e. (y »/»)"• = (y™) 1 '". To prove (b), observe that {(y 1/m ) 1/ "} m " = {{(y Vm y /n } n } m = (y" m ) m = y and hence (y Um ) Vn is the unique positive solution of x mn = y. To prove (c), observe that {y"» z >•'"}" = (ytoftfi-f = yz and hence y Vn z vn is the unique positive solution of x" =yz. We may now prove (iii), (ii) and (i) in the general case when r = pjm and x = qjn (p and q integers). (iii) yz r = ry)" m (z p ) ,/m = ov) 1 "" = &zff m = (yzf' m = (yzf 00 Iff = {{(y p ) 1/m ]°} 1/ " = {l(y p )"} Vm } l,n = lyP-y""" _ „pq/mn _ ,JS8 — > — >" (J) y r+s = y(P" + f")/mn _ r„pn qmil/mn _ (ypnyimnf qm\Vmn = yP""y<"" = y^ 4 Write ax 2 + bx + c = a(x - a)(x - fi).Whena< x <0,jc -a>0 and x-p<0. Hence (x - a)(x - (3) < 0. When .v < a, x - a < and x - (3 < 0. and therefore (a: - a)(x - 0) > 0. When *>/3,x-a>0 and x — > and therefore (* — a)(x — (3) > 0. 'Complete the square' as in § 1.10 to obtain ax 2 + bx + c = {(2ax + b) 2 - (b 2 - 4ac)}/4a. Since (2ax + b) 2 >0, Solutions to exercises (1.12, 1.20) 1 75 ax 2 + bx + c>c -\b 2 a' x with equality when lax + b = 0, i.e. x = —bjla. 5 Apply the Cauchy-Schwarz inequality using the numbers \Ja\,\Ja 2 , . . . ,\Ja n and 1/Va,, . - • , l/y/a„. We obtain n 2 = I y/ai . —r- + ■ ■ ■ + V<* n • -j— <( ai +a 2 + ...+a n ) -+J+... + - \a, a 2 and the result follows. 6 t kk + Kf = I 4 + 2 Z a h b k + I 6? fe = i k=i fe=i fc=i 1/2 / „ \l/2 n n n fc=i \fc=i / \fc=i / fe=i (Cauchy-Schwarz inequality) LI + U.«) Consider, for example, the case n = 3 and suppose that (xj,x 2 ,x 3 ), (y lt y 2 ,y 3 ) and (z h z 2 ,z 3 ) are the co-ordinates of the vertices of a tri- angle in three dimensional space. The length of the side joining the firs', two vertices is given by \l/2 fe=i < £ (x k -z k +z k -y k ? fe=i 3 fe = l 1/2 1/2 [,' U/2 • Exercise 1.20 1 We first show that \a\ < b implies -b < a < b and then that — fc < a < b implies \a\ < 6. (i) Suppose that \a\ < b. From theorem 1 .1 5, a < \a\ and \a\ >—a. Hence a <6 and — a< b, i.e. a >— 6. It follows that -fc< a < 6. (ii) Suppose that -b<a<b. Then a < b and -a < 6. Since, for each a, |a| = a or \a\ = —a, it follows that \a\ < b. 2 From theorem 1.18, 1 76 Solutions to exercises (1.20, 2.10) \c-d\>\c\~\d\ and \c-d\ = \d-c\>\d\-\c\ = -{|c| I Idl}. Thus, - {kl - \d\) < k - d\ < {kl - \d\} and the result follows from exercise 1.20(1). 3 Only (iv) is not immediate. We have d(x,y) = \x—y\ = \x — z + z — y\ < \x — z\ + \z — y\ = d(x,z) + d(z,y). 4 Suppose that r + s\/2 = t, where t is rational. Then, provided s ¥= 0, vi - £=. 5 But the right hand side of this equation is a rational number (why?) and so we have a contradiction. 5 We are given that a = r + s\/2 satisfies the equation ax 2 + bx + c = 0. Hence a(r + sy/2) 2 +b(r + s\/2) + c = {ar 2 + 2as 2 + br + c} + {2a + b}s^2 = 0. Since a,b,c,s and r are rational, it follows that (2a + b)s = 0. Thus a(r - sy/2) 2 + b(r - sy/2) + c = {ar 2 + 2as 2 + br + c) -(2a + b)s\/2 = and so & = r—s\/2 is also a root of the equation. 6 Suppose that m 2 = 3n 2 . Then m 2 is divisible by 3 and hence m is divisible by 3. (Try m = 2k + 1 or m = 3k + 2.) Hence m =■ 3k. But then 9& 2 = 3n 2 , i.e. 2k 2 = n 2 and so n is also divisible by 3. Suppose that m 3 = 2n 3 . Then m 3 is divisible by 2 and hence m is divisible by 2. (Try w = 2k + 1 .) Hence m = 2&. But then 8A: 3 = 2n\ i.e. 4£ 3 = « 3 and so n is also divisible by 2. • Exercise 2.10 1 (i) False (ii) true (iii) true (iv) false (v) true. 2 By exercise 1.20(1), || -jc|<5 if and only if -5 <£ -x<8. This last inequality is equivalent to 5 >x — £ > — 6 which is, in turn, Solutions to exercises (2.10, 2.13) 1 77 equivalent to % + 5 >x > g -5. But ft - 5, % + 5) = {x:£-5<;c<£ + 6}. 3 (i) (0, 1): bounded above; some upper bounds are 100, 2 and 1 , the smallest upper bound is 1 ; no maximum. (ii) (— °°, 2] : bounded above; some upper bounds are 100, 3 and 2; the smallest upper bound is 2 and this is also a maximum. (iii) {—1,0, 2, 5}: bounded above; some upper bounds are 100,6 and 5; the smallest upper bound is 5 and this is also a maximum. (iv) (3, °°): unbounded above. (v) [0,1]: bounded above; some upper bounds are 100, 2 and 1 ; the smallest upper bound is 1 and this is also a maximum. 4 (i) (0, 1): bounded below; some lower bounds are — 10, - 1, and 0; the largest lower bound is 0; no minimum. (ii) (— <*>, 2\. unbounded below. (iii) {—1,0,2, 5}: bounded below; some lower bounds are —10, —2 and — 1 ; the largest lower bound is — 1 and this is also a minimum. (iv) (3, °°): bounded below; some lower bounds are — 1 0, 2 and 3 ; the largest lower bound is 3; no minimum. (v) [0, 1] : bounded below; some lower bounds are — 10,-1 and 0; largest lower bound is and this is also a minimum. 5 {3,4}. 6 Take y = \x. No m e (0, °°) can be a minimum because \m is a smaller element of the set. \ • Exercise 2.13 1 Since B = sup 5 is an upper bound for S,x<B for each x G S. But S C5 means that, for each x G S , it is true that x^S. Hence x < B for each x G S . Thus B is an upper bound for S and therefore at least as large as the smallest upper bound sup S , i.e. sup S < B = sup S. 2 The proof is very similar to that of theorem 2.12. Let B = sup S and let T = {x + % : x G S). Since x < B for any x G S, it is true that % + x < \ + B for any x G S and hence that % + B is an upper bound for T. If C is the smallest upper bound for T, it follows that C<%+B. On the other hand, y < C for any y G T and therefore y — | < C— % 1 78 Solutions to exercises (2.13) for any y G T. Since S = {y — £ :y G 7"}, it follows that C — £ is an upper bound for S and therefore that B < C — £. We have shown that C< £ + B and O £ + 5. Hence C = £ + tf . 3 Let B = sup 5 and let 7" = {— x : x G 5}. Since x < 5 for any x G 5 it is true that — x > —B for any x G S and hence that — B is a lower bound for T. If C is the largest lower bound for T, it follows that C > —B. On the other hand, y ~5* C for any y G 7" and therefore — j> < — C for any 7 G T. Since 5 = {— y :y&T), it follows that — C is an upper bound for S and therefore that —OB. We have shown that C > -B and C<—B. Hence C = -5. To obtain the required analogues, consider a non-empty set 7" which is bounded below and then apply theorem 2.12 and exercises 2.13(1) and (2) to the set S = {- x : x G T). We obtain (i) inf $x = | inf x (£>0) i£r x e t (ii) If J C 7\ then inf T >MT (iii) sup (— x) = — inf x. x&T xGT 4 (i)l (ii)2 (iii) 1 (iv) 0. 5 (i) The set /J = {|| -x| :x G 5} has as a lower bound. If £ G 5, then is a minimum for D. Exercise 4(iv) provides an example for which tf(£,S) = Obutt£S. (ii) Let £ = sup S. Then |£ -x| = % -x for each x G 5. We therefore have to show that no ft > is a lower bound for the set D = {£— x :xGS}. If this is false, we can find an h > such that % -x > h for all x G S. But then x < % - h for all x G S and hence £ — ft is an upper bound for S smaller than the smallest upper bound. If £ = inf 5, consider instead d{~ I, T) where T= {-x :x G S}. (iii) Since / is an interval, | $ / implies that £ is either an upper bound or a lower bound for /. Suppose the former. Let B be the smallest upper bound of/. Then B G / because / is closed. Given any x G /, ||— #| = %~x = S-B+B-x = $-B+\B-x\. Hence, by exercise 2.13(2), inf »— x| = £-fi+ inf \B-x\ x e s % * 4 s and therefore d(|, 5) = £ - £ + d(B, S). But £ - B > 0, t/(Z?, S) > and d(t, S) = 0. It follows that £ = 5 and hence £ G /. Similarly if £ is a lower bound for /. Solutions to exercises (2.13, 3.6) 179 If/ is an open interval other than R or then £ can be taken as its supremum or infimum (at least one of these must exist). 6 We may assume that the sets S and Thave no elements in common (otherwise the problem is trivial). With reference to the given 'hint', the set To is not empty because t G T . The set T is bounded below by s. Let ft = inf TV ■f I huh m il i * To If ft ^ T, then ft G S. But ft is at zero distance from T by exercise 2.13(5ii) and hence we have found a point of S at zero distance from T. If ft G T, then ft > s and the interval (x, ft) is a non-empty subset of 5. Hence ft is a point of T at zero distance from S. • Exercise 3.6 1 We have n n for all n G (^ and therefore 1 is an upper bound for S. Suppose that 1 — ft, where ft > 0, is a smaller upper bound. Then, for all n G f^, l--<l-ft n n 1 -\ ' ?< ft and hence ft" 1 is an upper bound for N. This is a contradiction and therefore 1 is the smallest upper bound for S. The set S has no maximum. For no n G [^ is it true that h-1 = 1. 2 Suppose that S is bounded above. Then it has a smallest upper bound B. Since BX' 1 cannot be an upper bound, there exists an n G N such that X">BX' 1 X"* l >B. 180 Solutions to exercises (3.6) Hence B is not an upper bound for >*> and we have a contradiction. It is clear the is a lower bound for T. Suppose that h is a larger lower bound (i.e. h > 0). Then, for any « S fsj, x n >h l\" 1 It follows that the set S = {(llx)":n G M} is bounded above by h' 1 which contradicts the first part of the question. 3 Let S be a non-empty set of integers which is bounded above by B. Since F*J is unbounded above, we can find an n G N such that n> B. The set T~ {n— x:x£ S} is a non-empty set of natural numbers and hence has a minimum m (theorem 3.5). The integer n — m is then the maximum of the set S. Similarly if S is bounded below. 4 Since N is unbounded above, there exists a natural number n such that n>(b —a)" 1 . Otherwise (b —a)~' would be an upper bound for f^. Let m be the smallest integer satisfying m>an. Then m — 1 < an. Hence I m m 1 a < — and — < a + — < a n n n (b-a) = b as required. 5 Suppose thatm &S. Then \{tn + I)G5and^(m + \)>m. Hence m cannot be a maximum. Similarly m cannot be a minimum. It is obvious that S is bounded above by 1 . Suppose that H (0 < H< 1) were a smaller upper bound. By exercise 3.6(4) we could find a rational number r such that H<r<\. But then r would be an element of S larger than//. Thus 1 is the smallest upper bound. A similar argument shows that is the largest lower bound. 6 We know that all numbers of the form r\j2 are irrational provided r is rational and non-zero (exercise 1 .20(4)). By exercise 3.16(4) we can find a rational number r such that and hence a < r\j2 < b. (If r happens to be zero, locate a second rational number s satisfying <s\J7 < b.) Solutions to exercises (3.11) Exercise 3.11 181 1 (i)Lets„= l 2 + 2 2 '? + . . . + n 2 and let P(n) be the proposition that s„'= ««(« + 0(2n + 1)- Theni'(l) is true because s, = 1 and £l(l + 1)(2 + 1) = 1 . We next assume that P(n) is true and try and deduce thatP^i + 1) is true. s n + , = l 2 +2 2 +... + n 2 +(n+l) 2 = s n + (n+l) 2 = \n(n + l)(2n + 1) + (n + l) 2 (assuming P(n) true) = K» + l){«(2n + 1) + 6(n + 1)} = l(n + 1)(2« 2 + In + 6) = l(n + 1)(« + 2)(2« + 3) = l 6 m(m + l)(2m + 1) (m=n + 1). (ii) Let s n = l 3 + 2 3 + . . . + n 3 and let P(n) be the proposition that s n = kn 2 (n+ l^.ThenPO^s true because s, = 1 andjl 2 (l + 1) 2 = 1. We next assume that P(n) is true and try and deduce that P(n + 1) is true. s n +i = l 3 + 2 3 +... + « 3 + («+l) 3 = s„ + («+l) 3 = \n 2 (n + l) 2 + (n + 1 ) 3 (assuming P(n) true) = \(n + 1)V + 4n + 4) = |(« + \)\n + 2) 2 . 2 Let P(n) be the assertion 1 +x + ...+x" = 1 -x Provided x ¥= 1 , this is clearly true when n = 0. Assuming P(n) is true, 1 +x+x 2 + .. . + x n +x n * , = 1 -x n + l 1 -x l-x" + , +x" + 1 -x r \ l-x 1 -x n + 2 \-x and therefore P(n + 1) is true. The second formula of the question is obtained by writing a- =a/b. (Note that, if a = b or b = 0, the formula is obviously true.) 3 Suppose that P(!-) = 0. Then P(x) = P(x)-PQ) = a n {x n -% n ) + a n . l {x"- i -% n - i )+ ...+ ai {x-%). 182 Solutions to exercises (3.11) By the previous question we have x*-r = (*-£>(•* fe-l -L „&-2 s l-fc-2 J. f-fe-1 + r - l) where F k _ t (x) is a polynomial of degree A: — 1 . It follows that />(*) = (* -©WW*) +« B - 1 F n _ a €>c) + • • • + «,} = (a- -$)8(x). We apply this result to the second part of the question and obtain, in the first place, P(x) = (x — li)2i(x) where Qi(x) is of degree n — 1 . Then = P(% 2 ) = (£2 - f OGi(h). Since ? 2 # | b it follows that Qi{% 2 ) = and hence that Q t (x) = (x - % 2 )Q 2 (x) where Q 2 (*) is of degree n — 2. ThusP(x) = (x — £i)(x — % 2 )Q 2 (x). Next consider £ 3 and so on. B! ■ + - «! r-lJ r!(n-r)! (r- !)!(« -r + 1)! H! (r-l)!(n-r)! Ir n-r+1 £ 1 n\ n + 1 (r-l)!(n-r)! r(n-r+ 1) fe + 1)1 /« + l\ r!(«-r+ 1)! Let P{n) be the assertion that (a + b)"= I \ \a- r b r . r=0 Clearly P(l) is true. Assuming P(n) is true, (a+Z>)" +1 = (a + b)(a + b) n = a z (r=0 a b + b = £ f a »-v+ i(V~ r * r+I - = n\>- r = s:K'»' We replace r by s in the first sum and replace r + 1 by s in the second sum. Then (a + b) n * 1 = £ I U""' +1 6'+ I "J*""** 1 * Solutions to exercises (3.11, 4.6) 183 = a" z li 1 /« + 1 = z s = + , si \s-l n + 1 -slS _n+I -sts 6* + 6" as required. 5 Suppose there exists a natural number k for which P(k) is false. Since the set {2" : n e M} is unbounded above, we can find an M = 2 N > k. Now put S = {n : i>(") is false and n < M). This is a non-empty set of natural numbers which is bounded above. From exercise 3.6(3) it has a maximum element m. The assertion P(n) is true for m < n < M and hence P(m + 1) is true. But P(m + 1) implies P(m) and hence we have a contradiction. 6 From exercise 3.11(2), a "- b " = (a-b)(a"- 1 + a n - 2 b + ...+ab"- 2 Hence, if a >b >0, a" -6" fr"" 1 ). nb" < a-b <na" Therefore, If x > y, the result follows on writing a = x 1 '" and b =y 1/n . If .v<^,writea=>' ,/ "and6 =x Vn . • Exercise 4.6 1 Let e > be given. We must find a value of A' such that, for any n >N, n 2 -\ n 2 +l But « 2 -l -1 -1 <e. = 1 - n 2 -\ n 2 +\ n 2 +\ n 2 + 1 and so we simply need to find an N such that, for any n>N, ,7TT <e - i.e. \ 184 Solutions to exercises (4.6, 4.20) n 2 + 1 > 2/e. Choose N = (2/e) 1/2 . Then, for any n > N, n 2 + \>n 2 >N 2 = 2/e. We have shown that, given any e > 0, we can find an /V (namely (2/e) 1/2 ) such that, for any n >N, n 2 -\ n 2 +l Thus lim -1 <e. n 2 -l n 2 +l = 1. 2 Let e > be given. We must find a value of TV such that, for any n >N, 7" Q <e i.e. « r >l/e. We choose N = (1 /e) 1/r . Then, for any « > TV, n r >N r = 1/e and the result follows. 3 Let e > be given. We must find a value of N such that, for any n > N, \\x n -\l\<e. If X = there is nothing to prove. We may therefore assume X # 0. We are given that x n -* I as n -> °°. Since e/|X| > it follows that we can find an N such that, for any n>N, \x n ~l\<el\X\ i.e. |Xx„-X/|<e. This completes the proof. • Exercise 4.20 n 3 + 5n 2 + 2 1 + 5/T 1 + 2n~ 3 1+0 + 1 - as n •* °°. 2n s + 9 2 + 9n" 3 2 + 2 First consider the case when |x|< 1 . Then x" •* as n -><». Hence Solutions to exercises (4.20 J x+x" x + 1 +x"^ 1 +0 185 as n Next suppose that x > 1 . Then x " ■+ as n -*■ °°. Hence x+x" x.x~" + l + 1 asn 1 +x n x'"+l 0+1 This leaves x = 1 and x = —\ giving 1 + 1" = 1 and -1 +(-!)" 1 + 1" * 1 +(-l) n ' The latter expression is not even defined for odd n. Hence 1 (x> 1) x (— 1 < jc < 1 ) 1 (*<-!). lim x+x" ■»\l+x n limit undefined 3 We have 0<V(«+ i)-V« = (V(n+I)-V")(V(n + l)+Vn) (V(«+l)+V«) n+l-n 1 as« -*-°° V(«+l) + \/n V" (see exercise 4.6(2)). It follows from the sandwich theorem that V(h + 1) — V" "* as n •* °°. 4 Let N be the smallest natural number such thatN>x. Then, for n>N, x" X X X X X n~\ " T'2'N—l'N'"N X ,v-i / \n-N+l X IX i! I 2 N-l N Since x/N < 1 , (x/N)" -* as « -* °° and the result follows from the sandwich theorem. 186 Solutions to exercises (4.20, 4.29 "J 5 Suppose that (1 + l/») a+1 1*| > 1 for all n e N. Then n \x\ for all « S PiJ and this is a contradiction (see example 3.4). Hence, for some We W, (1 +UN) a + 1 \x\<l. If x ¥= 0, consider the expression a+l (n+lf +1 x iOr-H„n + l n a+1 x n I n 1*1. Un>N, \(n + l) a * l x n + 1 \<:\n a * 1 x n \. It follows that, for n>N, \n a+i x n \<\N a+1 x n \. We conclude that, for n>N, 1/iVK-UV' « Q+l^jVl and thus n°x" ■* as n -» °° by the sandwich theorem. 6 We wish to prove that This inequality is equivalent to (n + \) n <n n * 1 i.e. But we know from example 4.19 that (i) i + <3 for all n € H, Hence (1) holds provided 3 < n. • Exercise 4.29 1 (i) Let e > be given. We have to find an N such that, for any n>N, ||x„-/|-0|<e. Solutions to exercises (4.29) 187 But \\x n — /| — 0| = \x„ — 1\ and so what has to be proved is just the definition of x n -* / as n -* °°. (ii) By exercise 1.20(2), ||.x n |-|/ll<U„-/| and hence \x n \ -*■ |/| as n ■* °° by the sandwich theorem. 2 For any e > 0, there exists an N such that, for any n>N, l* B -l|<e. i.e. /-e<^:„</ + e. Since this is true for any e > 0, then it is true when e ■ "»|l (for an appropriate value of N, say W = A^). For « > TV, we therefore have that x n >l~\l - R 3 (i) Let // > be given. We have to find an N such that, for any n>N, T>H. Since the set {2" : « G M} is unbounded above (exercise 3.6(2)), there exists an JV6 N such that 2 N >H. If n >N, then 2">2 N >H and the result follows. (Alternatively one may appeal to the inequality 2" > n {n > 1) which is easily proved by induction.) (ii) Let H > be given. We have to find an N such that, for any n>N, -s/n<-H i.e. y/n>H i.e. n>H 2 . We may therefore take N = H 2 . (iii) The sequence <(— l)"n) cannot converge because it is unbounded. It cannot diverge to + °° because its odd terms are all negative. It can- not diverge to —°° because its even terms are all positive. 4 (i) Suppose that x n •* as n ■* °°. Let H > be given. We have to find an N such that, for any n >N, \/x n >H. But H' 1 > 0. Since x n -* as n -*■ °°, we can therefore find an N such 1 88 Solu [ions to exercises (4. 29, 5. 7) that, for any n>N, x n = fx.-OKJT 1 i.e. \lx n >H. (ii) To prove that \/x n -* + °° as n -* °° implies that x n -*■ as « -> °°, one simply has to reverse the above argument. 5 Let H > be given. If Cx„) is unbounded above, there exists an N such that x N > #. If (x n ) increases, then, for any n >N,x„ >x N > H. It follows that x n -*■ + oo. Similarly in the case when 0c„) decreases and is bounded below. 6 Suppose there is a value of n GN for which no x G S can be found satisfying |£ — x\< 1/h. Then 1/n is a lower bound for the set D = {\%—x\:xGS} which contradicts the assertion that dQ, S) = 0. That x„ -* i- as w -»■ °° follows from || — jc„| < l/« (« = 1 , 2, . . .) be- cause of the sandwich theorem. If £ = sup S, then from exercise 2.13(5ii) rf(|, 5) = 0. It therefore follows from the first part of the question that a sequence Cc n > of points of S can be found such thatJ£ n -» £ as « -» °°. (Note that the terms of this sequence are not necessarily distinct.) If S is unbounded above, then, given any n GH, we can find an x n G S such that* n > n. Hence x n -> + °° as n -> °°. • Exercise 5.7 1 Suppose that « 1/n -> / as n ■* °°. It follows from theorem 5.2 that (2rt) 1/2n -<-/asH-°°. But 2" 2n ■* 1 as n -*■ °° (example 4.14). Thus n v2 " -> / as n -*■ °°. Hence It follows that / = P. Since / > 1 , we deduce that / = 1 . 2 We prove that a<x n < bby induction. It is given that a < x 1 < b. If we assume that a < x n < b, it follows that x n+1 — a = x% + k—a>a 2 — a + k = 0. Similarly x n + s -b =xl+k-b<b 2 -b + k = 0. Hencea<x n+ , < b. (Note that the condition < k <$ ensures that the quadratic equation x 2 — x + k = has two real roots and that these are both positive.) Solutions to exercises (5. 7) 189 Next consider x n + l x n = x n X„ + k. But*?,— x„+k< because a <x n < b (see exercise 1.12(4)). This completes the proof that a < x n + 1 <x n < b. Since Cc„> decreases and is bounded below by a, it converges. Suppose that x n -* / as n -+ °°. Thenx n + 1 ->-/as« ^-°°. Butx n + i = x„ + k and so I = I 2 + k. This implies that l = aotl = b. But b cannot be the infimum of the sequence (x„> because it is not a lower bound. Hence / = a. 3 It is easily shown by induction that k > x„ > (n = 1 , 2 , . . .). Observe that l n + l X n -1 — l+X n l+*„- 2 k(X„-2 x n) Hence x n+i —x n . 1 has the opposite sign tox„.. 2 — x n . It follows, using an induction argument that one of the sequences <x 2n ) and <x 2n _j> increases and the other decreases. (In fact (x 2n -i) increases ifx 3 >*i and decreases if -X 3 < x t .) That both sequences converge follow from theorem 4.17. Suppose that x 2n •* I as n -*■ °° and that x 2n - 1 "* W as n -* °°. Then k 1 = k 1 +m' m l+l I + Im = k; m + Im = k. It follows that / = m and that I 2 + / = k as required. The conclusion about Ct,,) is that x n -*■ I as n -> °°. 4 If < a < 6, it is easily seen that the geometric mean G = V(a&) (see example 3.10) and the harmonic mean//= {\(\ja + 1/6)}" 1 (see exercise 1.12(5)) satisfy a<H<G <b. f We are given h= x i<yi= 1. On the assumption that x n _, <^ n _ 1 , x n-l< x n<y n -l because x n is the geometric mean ofx n _, andj'n-!. Further 190 Solutions to exercises (5. 7) x n <y n <y n -i because y n is the harmonic mean of x n and y n -j. It follows by induc- tion that x„_i <x n <y n <y„. l (" = 2, 3, . . .). The sequence (x n ) increases and is bounded above by y% — 1. The sequence (y n ) decreases and is bounded below by Xi = \. Hence both sequences converge. Suppose that x n -* I as n -* °° and y n -> m as n -> ° Then I 2 = Im m 2 \l m / Both of these equations yield I = m. 5 As in example 4.19 we apply the inequality of the geometric and arith- metic means with ai — Ot — . . .— fl„-i = (1 +>'/(" — 0) an d a n = 1 • Then . (n-l)/n II y n-\ < (»-l)(l+K« -!)-')+! _ L + Z\ n \ n Hence £T4+i This analysis is only valid if a u a 2 , ■ ■ . , a n are non-negative numbers and hence we have only shown that the sequence increases when 1 + y >0 i.e.n>\-y. n-l As in example 4.19, 1+ 4" <1+w+ a + ... + fi£. nj 2! n\ Since there exists an N such that, for any n>N, ff<6 (2) we may obtain from (2) iJV 1 _/■»•»" + ! <C+ 33— <C+2 where Cis a constant. r ♦■•■♦& Solutions to exercises (5. 7, 5.73^ i9/ It follows that -the sequence is convergent since it increases and is bounded above. Note that the limit is positive since 1 + y/n is positive when n is sufficiently large. 6 From the previous question we know that the sequence x 2 ] ~n 2 converges to a positive limit /. It follows that positive numbers a and b can be found such that for sufficiently large values of n. Hence «""<H"H" <4 '" The result then follows from the sandwich theorem and example 4.14. • Exercise 5. 15 1 For the sequence <(- 1)"(1 + l/«)> the set L = {- 1, 1}. Observe that (-l) 2n (l + l/2n)-> 1 as >?->•«• (_1) 2 " -1 (1 + l/(2«-l ))-*•-! as«-*-°° and the sequence possesses no subsequences which tend to a limit other than 1 or — 1 . It follows that (i) limsup(-l)"(l + l/«) = 1 00 lim inf (- 1)"(1 + l/«) = -1. Note that sup (-1)"(1 + 1/n) = (-1) 2 (1 +i) = I n > 1 inf (-1)"(1 + 1/n) = (-l) l (l + 1/0 = -2. n > 1 2 Observe that every rational number in the interval (0, 1) occurs infinite- ly often as a term in the given sequence. Let x £ [0, 1 ] . By exercise 3.6(4), a term /•„_ of the sequence can be found such that x-\<r„<x+ 1. A term r„, with n 2 >«i can then be found such that 192 Solutions to exercises (5.151 x-\<r n <x + \. Continuing in this way we construct a subsequence <r„ fc > such that and hence r„ fc -* x as k •* °° by the sandwich theorem. 3 The hypothesis of the question implies that Ce n ) contains a subsequence all of whose terms are at least as large as b. By the Bolzano- Weierstrass theorem, this subsequence in turn contains a convergent subsequence <x„ r >. Suppose that x„ r -* I as r ■+ °°. Since x„ r > b (r = 1 , 2, . . .), it follows from theorem 4.23 that / > b. 4 Suppose it is false that, for any e > 0, there exists an N such that, for any n >jV, x„ < /"+ e. Then for some e > it is true that for each A' we can find an n>N such that x n > 1 + e. From exercise 5.15(3) we can then deduce the existence of a convergent subsequence with limit / > /"+ e. This contradicts the definition of /"(see §5.12). The corresponding result for Hs that for any e > 0, we can find an A' such that, for any n >N,x n >/ — e. 5 Suppose that [ = 1 = 1. Let e > be given. By the previous question we can find an A'i such that, for any n>N t , Solutions to exercises (5.15, 5.21) 193 (3) x n < I + e. We can find an N 2 such that, for any n>N 2 , l-e<x n . (4) Take N = max {#,, N 2 }. Provided n > N, inequalities (3) and (4) are true simultaneously. Hence, for any n >N, I - e < x n < I + e i.e. |x„-/|<e. Thus x n -> / as n -* °°. That x n -*■ I as n -*■ °° implies I = 1 = 1 is trivial. To prove the last part of the question one only has to note that, if m is the limit of a convergent subsequence, then / < m < I. The sequence <M n ) decreases because we are taking the supremum at each stage of a smaller and smaller set. Any lower bound for Cc n > is also a lower bound for <Af„>. It follows from theorem 4.17 that <M„> con- verges. Suppose that M n -+M as n -*■ °°. Suppose that x n -*lasr-+°°. Then, for n T > n. x nr <M n and hence KM n by theorem 4.23. From theorem 4.23 again it follows that / < M . This is true for each / e L and so /"< M. From exercise 5.1 5(4) we know that, given any e > 0, we can find an n such that, for any k > n, x k < 1+ e. Thus /"+ e is an upper bound for the set {x k : k > n} and it follows that M<M n <l + e. Since this is true for any e> 0, M < /"(see example 1 .7). We conclude that M = / i.e. lim I sup x k \ = Urn sup x n . n->°°\h > n ) n -* « • Exercise 5.21 1 If«>m, \x n —x m \ < \x n - X„-i I + \x n - 1 -x n . 2 \ + . . . + \x m+l -x m \ <,a n - 1 + a"' 7 + . .. + a m = a" 1-a" 1 -a < 1-a Given e > 0, choose N so large that, for any m>N, _m <e (Example 4.12). 1 -a The result then follows. Exercise 4.20(3) provides the example y n = \/n (n = 1 , 2, . . .). 2 Note first that a < x n < b (n = 1 , 2, . . .). Hence a ^. x n + 1 b b x n a Next observe that x n + 2~ x n + l ~ X n + \X n —X n+i = JC n + i(X n — X n + i) X n + l \X n —X r \X n *2 X n + i \ — |A„ A n + n Xn+2 ' Xn+l < a + b \X n -Xn+ll It follows that 194 Solutions to exercises (5.21) , \n-l |x n + 1 -*„|< a + b \X 2 — Xy\ and so we may proceed as in the previous question. i It x n +2 = %(X n + x n + i), X n + 2 <" 2~X n + i = X n+ ! + 2~X n = . . . = X2 + jXj. Put l = l(x2 + kxt). Then *n + l "*" 2*n = 2' •^n + l-^ = l(/-^ n ). Hence l*n + i-/l = \\x n -l\ =...= — |*i - /| ->0 as h -*■■». It follows that x n ->■ / as w -»• °°. '* -^n + 2 = i*n + l*n/ > Xn + 2 x n+l ~ X n + i X n = . . . = X 2 Xi. Put /={*2*.}" 3 - Then and it follows that x n -*■ I as n -* °°. 4 Let <x n > be a sequence of points in [a, Z>]. Since [a, fcj is bounded, the sequence <*„> is bounded. By the Bolzano-Weierstrass theorem, the sequence therefore has a convergent subsequence (x n >. Suppose that Xn r -* / as r -* 00. Since a < *„ r < 6, it follows from theorem 4.23 that a < / < b. Hence (ae«> converges to a point of [a, 6] . 5 Suppose that /is unbounded above. By exercise 4.29(6), we can find a sequence <x n ) of points of / such that x n -*- + » as « -»• °°. But no sub- sequence of such a sequence can converge and so this is a contradiction. Hence / is bounded above. Let b = sup /. By theorem 4.29(6) we can find a sequence 0e„> of points of/ such that x n •* b as n «* °°. yl// sub- sequences of tx„> also converge to b (theorem 5.2). Hence be I. Similarly for a = inf /. 6 Let (x n ) be a sequence of distinct points of 5. Since (x n ) is bounded it contains a convergent subsequence (x n >. Suppose that x n ■* § as /■ ■* °°. Then £ is a cluster point of 5. Solutions to exercises (6.26 J 195 • Exercise 6.26 1 We have 3rc-2 n(n + l)(« + 2) It follows that & 3«-2 1 5 = + fft «(« + !)(« + 2) n m + 1 n + 2 = (-*+!-# + (-i+/-l) +Hf+X-A 2 A convergent sequence is bounded (theorem 4.25). It follows that, for some H, a n <Hb n (« = 1,2,...). Thus the convergence of S"= 1 &n implies that of S„ = , a„ by the comparison test. Since / > 0, from exercise 4.29(2), there exists an N such that, for any n > N, a n >\lb n and hence the convergence of 2™=i a n implies that of 2~=i ft n by the comparison test. (i) Take a n = — and b n = -. Since £ - diverges (theorem 6.5), it 2« « n=l" 00 j it follows that £ — diverges. n=l *" 196 Solutions to exercises (6.26) 1 1 °° 1 00 Take a n = — and b n = -. Again V - diverges. m — i n n - j ln — \ 2 1 °° 1 (iii) Take a n = 2 and b n = -5. Since £ -5 converges (theorem s 2 6.6), it follows that £ . converges. „=i 1+3 3 We have 2n n b n = a n+ , +.. . + a 2n = 2 a fc - J] <z fe ->- as h -* °°. Since the sequence (a„> decreases, b n >na 2n >0. It follows from the sandwich theorem that 2na 2n "* as n -> ■». Similarly (2« — l)a 2n ., -*• as n -* °°. 4 The series diverges because its terms do not tend to zero (see example 5.11). 5 (i) We use the ratio test. 'n*l = ((« + l)!) 2 (2n)! = (n + l) 2 1 ~ (2(/i + 1))! * («!) 2 (2« + 2)(2n + 1) "* 4 The series therefore converges. (ii) Again the ratio test may be applied. a n + l _ ((n+l)l )\ n+l (2n)\ 1 5 a n -^TTyy. 5 ^7?^ 4 as " "* °°- The series therefore diverges, (iii) We use the nth root test asn n n+ 1 1 + — I ) -» — as n -*■ °o. The series therefore converges because e > 1 . (iv) Again the nth root test may be applied. n ,n + 1 4" Vn — as n -*• °°. e The series therefore diverges because e < 4. Solutions to exercises (6.26, 7.16) 197 (v) We have V0L±JW« = I : _<1 („-l,2,...) n «(V(« + i) + V«) "' and hence the series converges by the comparison test (see theorem 6.6). (vi) This series converges by theorem 6.13. 6 We have *-»*£-B + H + - + 1 .+. ! 4fl - 3 4n - 1 2n = 1 + I + i-i + i 2 3 4 5 1 1 2 2n I-... + — 2« + 2 2« + 4 An = s 2n + - JL + _l_ + ...+i 1 . k + 1 « + 2 2«| Here s„ denotes the /ith partial sum of the given conditionally con- vergent series with sum s. Observe that 2 3 2n = 2 1 1 - + -■ 2 4 -H- 2k 1 2n From this identity it follows that 1 1 n+ 1 n+2 Hence f3n = S 2n + 2* 2n "* I s aS « + il Exercise 7.16 1 5 - 1 1 1 1 1 1 J T 1 1 1 I Y//////7\ 1 2 198 Solutions to exercises (7.1 6 J Each vertical line meets the graph in one and only one point. The range is {2, 5}. The image of the set [1,2] is {5}. (i) The equation does not define a function from R to itself. The value x = 2 is an example of an x for which no corresponding y exists. (ii) The equation does not define a function from [— 1, 1] to R. The value x = is an example of an x for which two corresponding values of y (namely + I and — 1) exist. (iii) The equation does define a function from [— 1, 1] to [0, 1]. For each x satisfying — 1 < x < 1 there is precisely one y with < y < 1 satisfying |xj + |.v| = 1, i.e.y = 1 - \x\. 3 We have /o<fr) - me* ~ \=& « ! Ttl 1 :3 (o<,<i) + g(x) l+4x(l-;t) 1 -x Ix go fix) = g(f(x)) = 4/(*)(l -/(*)) = 4 - \l +xj \l +x (0<x<\). The two formulae give different results when x = 1 and hence define different functions. In order that/" 1 : [0, 1] -* [0, 1] exist, it must be true that, for each y satisfying < y < 1 , the equation l-x has a unique solution x satisfying < x < 1 . This requirement is easily checked by solving the equation y +yx = 1 —x -y x = l+y Solutions to exercises (7.16) The inverse function therefore exists and 1-v 199 r'o) = (0<y<\). l+y (Note that / = r X )- The equation y = 4x(l -x) does not have a unique solution x satisfying < x < 1 for each y satisfying < y < 1 . For example, 44(1-1) = 4.|(1 -f) = I. Thusg" 1 does not exist. 4 The function /has no inverse function. When^ = — 1 , for example, the equation^ = x 2 has no real solution at all. The function g has an inverse function. For each>- > there exists precisely one x > such that y = x 2 which we denote by y/y (see §1.9). Hence g' 1 (y) = y/y O>0). 5 We have j' =g(x)if and only ifx =g~ l (y). (i)g- 1 o g(x) = £-%(*)) - r'OO = x (xGA) (ii)g o g'Hy) = g(g~ l (y)) = g(x) = y 0' 6 B). When g is as in question 4, these formulae reduce to yjx 2 = x (x>0) (\/y) 2 = y (y>o). 6 (i) We apply exercise 2.13. Put 7= {f(x):x<= S}. Then sup {f{x) + c} = sup {y + c] iES y e T = c + sup y = c + sup f(x). (ii) Let H = sup fix); K = sup gix). x e s «6S Then, for all *e S, fix) + gix)<H + K 200 Solutions to exercises (7.16, 8.15) and hence H + K is an upper bound for the set {f(x)+g(x):x£S}. The result follows. To show that equality need not hold, take S = [0, 1] and let f(x) = x(0<x<l)andg(x) = 1 -x (0 <*< 1). Then f(x) + g(x) = 1 (0 < x < 1 ). Thus sup {/(*)+ #00} = 1 X es but sup /(*) + sup #00 =1+1=2. x es iss • Exercise 8.15 1 Since rational functions are continuous at every point at which they are defined (theorem 8.13), we obtain x 2 + 4 1+4 5 W Ai»H " 1-4 3- , . , ^ 73 + 5x 42 + 9 0+5.0 + 9 (ii) im t; = z v '*-o 3* 23 + 7 3.0 + 7 9 * 7 /(*) = 3-JC (jc> 1) 1 (* = 1) 7x (x<\). (0/00 + 2 as* 5 > such that l/W-2Ke 1 — . Given any e > 0, we must show how to find a provided that 1 — 6 < x < 1 . Since we are only concerned with values of* satisfying* < 1 , we can replace /(*) by 2x. The condition 1/00 — 2| < e then becomes \2x - 2|< e, i.e. 1 - |e < x < 1 + §e. We therefore have lo find a 5 > such that he 1 - \e < x < 1 provided that 1 — 6 < x < 1 . The choice 6 = ie clearly suffices. Solutions to exercises (8. 15) 201 (ii)/00 ->■ 2 as x -* 1 +. Given any e > 0, we must show how to find a 5 > such that 1/00 -2|<e provided that 1 < x < 1 + 5. Since we are only concerned with values of* satisfying* > 1, we can replace /(*) by 3 — *. The condition 1/00 -2|<e then becomes |3 -x — 2| <e, i.e. 1 -e<x<l + e. The choice 5 = e therefore suffices. It follows from proposition 8.4 that lim /OO X -» 1 exists and is equal to 2. Note that/(l) = 1 and so the function is not continuous at the point 1 . 3 lf**0, . . . l+2*+* 2 -l /OO = = 2 + x. Since in considering a limit as * ■+ we deliberately exclude considera- tion of the value of the function at * = 0, it follows that lim /OO = lim (2+*) = 2. * -» x -* 4 Let e > be given. We must find a 5 > such that |* — %\ < e provided that < |* — 51 < 5- The choice 5 = e clearly suffices. Observe that l/&)-|KlS-*l and hence /(*) •* % as * -> J by the sandwich theorem (proposition 8.14). 5 If£>0,let0<Jr<$< Y and take % =y in exercise 3. 11 (6). If X < x < Y, L Y 1 ' n Y- i \x-^\<\x 1 "'-^ /n \<-X lln X- 1 \x-^ n n and the sandwich theorem applies. To prove that /(*) -* as * -*■ +, we must show that, given any e > 0, we can find a 6 > such that x Vn m |/n_ | <£ provided that < * < 5 . The choice 5 = e" suffices. 202 Solutions to exercises (8.15, 8.20) 6 For any e > 0, we can find a 6 > such that |/(*)-/|<e provided < \x — £| < 5 . Since this is true for any e > 0, it is therefore true for e = I (with an appropriate value of 6, say 6 = h). We then obtain \f(x)-l\<l i.e.O = l-Kf(x)<l + l provided that < \x - §| < ft, i.e. §— h <x < £ + h and* ^ £. (i) if / < 0, an A > can be found such that /"(x) < provided that Z-h<x<H+handx*t. (ii) If / = 0, nothing of this sort can be said. • Exercise 8.20 1 (i) Given // > 0, we must find a 5 > such that x' x <-H (5) provided that -6 <x<0. If x <0, (5) is equivalent to 1 >-Hx which is in turn equivalent to -H' l <x and hence the choice 6 =H~ 1 will suffice. (ii) Given H > 0, we must find an X such that x 2 >H provided that x > X. The choice A" = yjH suffices. 2 Let e > be given. We must find an X such that |/fe(*))-/|<e provided that x > X. Since /(y) -* I as >> -* + °°, we can find a Y such that l/0)-/|<e (6) provided that y > Y. (7) Since g(x) x>X, g(x) > Y. 1 as x -* + °°, we can find an X such that, for any Solutions to exercises (8.20, 9.1 7) 203 If x > X, it follows that (7) is satisfied with y = g(x) and hence that (6) is satisfied with;- = g(x). But this is what we had to prove. The problems of theorem 8.17 arise because of the possibility that g(x) = r? for some values of x. In the above problem 77 is replaced by + °° which is not a possible value for g. 3 Let e > be given. We have to find an X such that, for any x > X, Since /(y) -> / as >> -*■ +, we can find a 5 > such that l/0)-/|<e provided that < y < 5. We choose X = 5~ l . Then, if x > X, y = X H <X~* = 5 and the result follows. 4 Suppose that f(x) ■* \ as x -* £. Then f(x n ) -*■ X as n -> °° and f(y n ) -*■ \ as n -» °° (theorem 8.9). This is a contradiction. 5 Takex n = 1/n (« = 1, 2, . . .) and;' n =y/2/n (n = 1, 2, . . .). Then /(x n ) = 1 -*■ 1 as n -» °° and /(y„) = -» as n -> «». Hence lim fix) x -* does not exist by the previous question. 6 We have < \xf(x)\ < \x\ and so the result follows from the sandwich theorem. • Exercise 9.17 1 113 — 2 2j + 3 k. ! i > J- = |.v - II Continuous on ( — 2, 2) and on [0, 1], Recall that a rational function is continuous wherever it is defined. Continuous on (—2, 2) and on [0, I ]. Note that the function has a 'corner' at x — 1 . 204 Solutions to exercises (9. 1 7) Solutions to exercises (9.17, 10.11) 205 ,l " 1 A> , 1 - 1 [l * (iv) Not continuous on (— 2, 2) and not continuous on [0, 1]. We have/tr) -* 1 as X — ■ 1 — but /(l) = 2. (v) Not continuous on (—2, 2) but continuous on [0,1]. j"-*±Jer#±a Continuous on (— 2, 2) and on [0,1]. Continuous on (— 2, 2) and on [0,1], All of the functions except (v) are bounded on (—2, 2): all but (i) and (ii) attain a maximum on (—2, 2): all but (i) and (v) attain a mini- mum on (—2, 2). 2 Let g: 1 -* R be defined by g(x) = f{x) — x 2 . Then g is continuous on /. Given any point g E I, let </•„> be a sequence of rational numbers in / such that r„ -* g as n -» °°. Then #(r„) ■* g(g) as n ■* °° (proposition 9.6). But g(r„) = (n = 1, 2, . . .). It follows that =*«) = /(g)-g 2 . 3 Let P be a polynomial of odd degree n. Let /•(jc) = a n x n +a n . l x n - l + ...+a 1 x+a a„ + + . . . + -=rr. + -= x x ' x lfa n >0, then P(x) -* + °° as x -* + °°andP(x)^ — °°asx -*■— °°. Hence we can find a and b s©h that P(a) < < P(b). Since P is con- tinuous everywhere, it follows from corollary 9.10 that there exists a g between a and b such that P(£) = 0. (Similarly if a n < 0.) 4 Suppose that g can be found with /(g) > 0. Since fix) -* as x -» + °°, we can find 6 > g such that, for any x > b, \f{x)\ < /(g). Also, since f(x) -*■ as x -* — °°, we can find a < g such that, for any x < a, \f(x)\ </(g). By theorem 9.12, /attains a maximum on [a, b]. Suppose that /(t?) > f(x) (x E [a, b ] ). Then /(t?) > /(g) > f(x) (x ^ [a, b]). Hence/attains a maximum on R. If a | can be found for which /(g) < 0, then a similar argument shows the existence of a minimum on R. If f(x) = for all x, the result is trivial. 5 Let Xj E I. Then there exists an x 2 E I such that |/(x 2 )l < |l/C*j)l and there exists an x 3 E I such that |/(x 3 )| < \\f(x 7 )\ and so on. We can therefore find a sequence (x n ) of points of / such that l/(*»)l < \ l/(*»-.)l < p 1/(^-2)1 < • • • < ^ l/fedl. It follows that /(x„) •* as m -* °°. Since / is compact, the sequence (x n ) contains a subsequence <je„ r > which converges to a point g E I (exercise 5.21(4)). Because /is continuous on /,/(*„,.)-* /(g) as r -* °° (prop- osition 9.6). Hence /(g) = 0. 6 That /is continuous on / follows from proposition 9.3. We have \x n + 1 -x n \ = \f(x n )-f(x n .i)\<a\x n -x n . 1 \. Hence |x B+1 -ayKaP-Haa -*il (« = 2, 3, . . .). As in exercise 5.21(1), we may conclude that (x n > is a Cauchy sequence and hence that 0c n > converges. Suppose that x n -*■ I as n -* ■». Because / is closed we have that / E I. But /is continuous on / and therefore we may deduce from the equation x n ►1 = f{Xn) that /=/(/)• • Exercise 10.11 1 We have to consider 1 1 ) 1 h 1 + (x + hf \+x 2 \ h 1 +x 2 -\ -x 2 -2xh-h 2 (1 + (x + h) 2 )(l +x 2 ) 206 Solu tions to exercises (10.11) -2x-h -2x (i + (x + hy)(i +x 2 ) (1+ x y as h -> 0. 2 Let n = —m where m£N. Then, by theorem 10.9(iii), 1 x m . - 1 . w*" Z). v " = D — = 3 lfft>0, /(1+/Q-/(1) _ 2 + 2/; -2 h lfh<0, f{\+h)-f{\) _ (1 +/!)'+ 1-2 = 2 -► 2 as A ->■ ■ ft ft = 2 + ft->-2asft->0-. That/'(1) = 2 then follows from proposition 8.4. On the other hand, if h > 0, g(0 + h)-g(0) m ft-0 ft ft but, if ft <0, *(0 + ft)-*(0) -A-0 -> 1 as ft -* + ft ft • — 1 as ft -> — . 4 By exercise 3.1 1(3), Pfr) = (x — g)R(x) where R is a polynomial of degree n — 1 . Thus /•'(*) = £(*) + (jc - %)R \x) (theorem 10.9(h)). Since P\%) = it follows that R (£) = and therefore /?(*) = (•* — Dfi(*) where Q is a polynomial of degree n — 2. 5 Put #,(*) = (*-?)'. Then flfeft) = /(/-I)... (/-ft + 1)(* -©'"" (ft = 0,1, 2,. ..,/)■ It follows that £*&(£) = (ft = 0, 2, ...,/- 1) and D l g,(x) = /! (xGR). Hence D k g,(x) = (ft = I + 1, / + 2, . . .). We now apply these results to the 'Taylor polynomial' P and obtain D k p$) = o + o + . . . + i /o^dz^cd + . . . + o = / <fc, <£) (ft = 0, !,...,« -1). Solutions to exercises (10.11. 10.15) 207 6 We prove the result by induction. It certainly holds for n— 1. If DJg = i h rffir-ig then D n + l fg = I ("] {D i+1 fD n - i g + D i fD n -^ 1 g} and the proof continues as in the solution to exercise 3.1 1(4). • Exercise 10.15 1 Let.y >x. Then, by exercise 3.1 1(6) withX = * and Y=y, i v I/n_ y l/n 1 It follows from the sandwich tlieorem that lim * — = -x^x~\ y-*x+ y— x n A similar argument shows that the left hand limit is the same. 2 Write r = m/n where m is an integer and n a natural number. Then D{x m,n } = D{(x m ) vn } = - (x m ) 1 l n x- m mx m - 1 - -x m/B *- 1 . 3 (i)Z){l +x" 27 } 3/5 = |{1 +x 1/2 T 2/S •^^" 26,27 (*>°)" (ii)5{x + (x+x 1,2 ) 1/2 } 1/2 = I {x + (x + x v Y 2 }- V2 { 1 + le* + *" 2 )~ 1/2 (i + K" 2 )} 6» > °)- 4 f {/(* 2 )} = /V)2x; f {/(*)} 2 = 2f(x)f\x). dx ax When x = 1 , 2/'(l) = 2/(l)/'(l) and the result follows. 5 By theorem 10.13, 1 = Dx = /)(/-' o /)(*) = Df-^Dfix). 208 Solutions to exercises (10.15, 11.8) y •v =£-'(>•) Observe that, for h > 0, (0 + /0 i/3_ i/3 = ^^ H Thus^" 1 is not differentiable at 0. °° as h -*■ +. 6 If* is rational, then y =f(x) = x is rational and so fo f{x) = /(/(*)) = /O) = j, = /(*) = *. If x is irrational, then^ = f(x) = —x is irrational and so fO f(x) = f(fix)) = /O) = -y = -fix) = *. A thoughtless application of theorem 10.13 would yield 1 Dfix) = Dfiy) (V =/(*))• But the function /is not differentiable at any point - thus nothing can be deduced from theorem 10.13. • Exercise 11.8 1 The stationary points are the roots of the equation /'(x) = 0. But fix) =x 3 -3x 2 + 2x and so fix) = 3x 2 -6x + 2 'y=x(x- l)(x-2) Solutions to exercises (11.8) 209 and hence the stationary points are given by 3±V(9-6) 1 x = 3 ^VS' If the maximum on the compact interval [0, 3] is not attained at an endpoint, then it must be achieved at a stationary point (theorem 1 1.2). The only possible candidates for a point at which a maximum is achieved are therefore x = 0,x = 1 ± 3" 1 ' 2 and* = 3. If /is evaluated at each of these points, one finds that/(3) = 6 gives the largest value and hence 6 is the maximum. A similar argument shows that — 2/3 \/3 is the minimum and this is achieved at x = 1 + 3~ 2 Choose the constant h so that F = f + hg satisfies F(ff) = Fib) and thus Rolle's theorem may be applied. We need fia) + hgia) = fib) + hgib) i.e. fib) -fid) gib)-gia)' We obtain the existence of a £ G (a, b) such that o = f\%) = fm+hg'm i.e. g'm and the result follows. h = -' 3 Apply the previous result with b = x. We deduce the existence of a % G (a, x) such that fa) m /(■*) g'm g&) ' Observe that % depends on the value of x, i.e. '£ is a function of x\ By exercise 8.1 5(4), we have % -*■ a as x -*■ a. Moreover %i= a when x > a and so it follows from theorem 8.17(ii), that ,. fi%) .. f(y) hm -7— = hm ,, . *-* a+ gii) y-*°*g(y) provided that the second limit exists. The result follows. lim ^~V(1 +y 2 ) = lim . y -►too"' * V " ' X->0+ \X '' + ? 21 Solu tions to exercises (11.8) = lim 1 -V(l +x 2 ) = lim = 0. x -* o+ 1 4 Given that /'(*) = x 2 for all x, it follows that D{f(x)-\x 3 } = f'(x)-x 2 = for all x and hence there exists a constant c such that fix) — %x 3 = c for all x (theorem 1 1 .7). 5 From Rolle's theorem we may deduce the existence of numbers J?o.n»i • • • ,V n -i such that | <T?o<^i<'7i<- ■ -<g n -i <*?„-, <g n and f\Vi)-P\m) = (i = 0, l,...,n-l). Containing in this way we can demonstrate the existence of a £ such that f M i%)-P {n \%) = 0. But a polynomial of degree « — 1 has the property that P^"\x) = for all*. 6 Since g(0) =g(i) = 0, it follows that/(0) =/(l) = 1 and so Rolle's theorem may be applied on the interval [0, 1]. We obtain the existence of a g £ (0, 1) such that /'(g) = 0. But then /(g) = 1 and so Rolle's theorem may be applied on the intervals [0, g] and [g, 1 ] and so on. This argument shows that between every pair of points x and .y for which f(x) — f(y) = 1 , we can find a point z for which /(z) = 1 . Suppose that < a < b < 1 and (a, b) contains no point x for which f{x) = 1 . Let A = sup{x:*<aand/(.x) = \}\B = inf {y :y>b and/O) = l}. Because a<b,A<B. Also, the continuity of /implies that fiA) =fiB) = 1 [use exercise 4.29(6) and theorem 8.9]. But we know that a point C can be found between A and B for which /(C) = 1 and this is a contradiction. Let* G [0, 1]. Since every open subinterval of [0, 1] contains a point g for which /(g) = 1, we can find a sequence g„ of points of [0, 1] with/(g n ) = 1 such that % n -*x as«^-°°. This implies that fix) = 1 because /is continuous. 1 Solutions to exercises (11.11) 211 • Exercise 11.11 1 Let /: U -> U be defined by fix) = x" and apply Taylor's theorem in the form f(y) = m)+Y^-^'^ + --- + h (y ' m ' n)m+En+v Put y = x + 1 and g = 1 . One then has to observe that, for /c = 0,l,.. .,«, n(n -!)■■■(« -* +l)g"- fe i ft! j^/ni) = and that 1 X n + 1 / (n + l) (7?) = 0. (" + 0! 2 By Taylor's theorem 0+4) 1 ' 2 = 2+^4-" 2 +^-Ki-i)4- 3 ' 2 +j ? -ia-i)a-2)'?- s/2 where 17 lies between 4 and 5. Hence But n 5/2 -v, 45/2 _ 2 s and the result follows. 3 By Taylor's theorem, for any xGl, fix) = /(g) + ^A?) + .-. + ^9"/ (,, - 1) «) + ^ Hence, with the given information, for some 7? between * and g. Since / (n> is continuous at g, it follows from exercise 8.15(6), that, if x is taken sufficiently close to g, then / (n) (7?) will have the same sign as/ (n, (g) (provided that/ (n) (g) ¥> 0). The problem therefore reduces to examining the sign of the right hand side of (8) under the various cases listed. (i)/ (n >(g) > and n even. Then the right hand side of (8) is positive for x sufficiently close to g and hence /has a local minimum at g. (8) 212 Solutions to exercises (11.11, 12.12) (ii)/ (,1> (£) > and « odd. If x is sufficiently close to f, then the right hand side of (8) is positive if x > £ and negative if x < %. (iii)/*"^!) < and n even. We obtain a local maximum. (iv)/ (n) (?) < and n odd. This case is similar to (ii) - see the diagram below. (i) Solutions to exercises (12.12) 213 (iii) (iv) Of the three functions given, (v) falls into class (ii), (vi) falls into class (i) and (vii) into class (iii). • Exercise 12.12 1 We have f(a) < f(x) < f(b) (a<x< b). 2 Observe that /'(*) = -(_x+]) ll "(x+\)- , --x l "'x- 1 <0 (x>0) n n because n > 1 and hence i+ £)H) >] (x>oy 3 If x and y lie in /and. y >x, then f(y) >f{x). Hence m= Wm m=M m lim /W^,o. y->* y— x y-**+ j — x The function/: R ■+ R defined by /(a:) = x 3 is strictly increasing on Rbut/'(0) = 0. 4 Since/is continuous, /(R) is an interval (theorem 9.9). Since f(x) -* + °° as x -» +«> and /(x) ■* - °° as x -*■ — °°, it follows that /(R) = R . We have fix) = 1 + 3x 2 >0 (*eR) and hence theorem 12.10 assures us of the existence of an inverse func- tion f' 1 : R •* R . lie unique value of x satisfying — \=y = f(x) = 1 + x + x is obviously x = — 1 . Hence 1 1 I Df(x) " £>/(-!) 4" DrH- 1) = Df-'iy) = 5 Suppose that % is not an endpoint of/. To show that /is continuous at £, we need to prove that/(|-) =/(?+) (see corollary 12.5). If /(£ -) < /(£ +) then we can find X such that /(£ -) < X </(£ +) and X =#/(!). Choose a and ft in / so that a < % < b. Then /(a) < X < f(b) and so we can find an x e / such that /(*) = X and hence f(X ~) < /(*) < /$ +)• Since f&) *f(%) this is in consistent with corollary 12.5 and hence /(£-) =/(£ +). A slightly different argument is required if % is an endpoint. The function /: (0, 2) ■* R defined by m = x (0<x<l) X-J (Kx<2) shows that the requirement that /be increasing cannot be abandoned (although it might be replaced by some new condition). For each x satisfying < X < 1 , < f(x) < Mix). It follows that /(*)' 0(x) = lim Mix) exists for each* satisfying <x < 1 and that 0(x) = or 0(x) = 1. If /increases on [0,1], then/(x) = Mix) for each x satisfying < x < 1 and hence 0(x) = 1 (0 <x < 1). Thus is continuous. It is more difficult to deduce that /is increasing given that is con- tinuous. We show to begin with that, if is continuous on [0, 1], then <fi(x)= 1 (G«X<1). We know that 0([O, 1 ]) C {0, 1}. But, by theorem 9.9, the image of an interval under a continuous function is another interval. Hence either 0([O, 1]) = {0} or 0([O, 1]) = {1}. But the former case is impossible because 0(0) = 1 . Thus 0(x) = 1 (0 < x < 1 ). Let < x <y < 1 . We know that fix) = Mix) and fiy) = M(y). Obviously Miy)>Mix) and it follows that / increases on [0,1]. 214 Solutions to exercises (12.21) • Exercise 12.21 1 We have f(x) = (x - l)(x - 2)(x - 3) = x 3 - 6x 2 + 1 lx - 6. Hence f'(x) = 3x 2 -12x + ll f"(x) = 6x-12 = 6(x-2). If follows immediately that / is convex for x > 2 and concave for x < 2. The roots of the quadratic equation 3.x 2 — 12x + 1 1 are x = 2± 3" 1/2 and it follows that /decreases on the interval [2 - 3~ l/2 , 2 + 3" 1 ' 2 ] ' and increases on the intervals (— °°, 2 — 3 1/2 ] and [2 + 3 -1/2 ")■ 2 Take*i = x,x 3 =y andx 2 = ax + /3y in (1) of §12.13. Since a + 0=l, x 2 = ax t +(1 -a)x 3 and a = x 3 -x 2 = x 2 ~Xx X 3 Xi X 3 X\ Thus (1) of § 12.13 is equivalent to (*3 -*l)/(*2) < (*3 -X 2 )f(X 1 ) + (X 2 -X l )f(x 3 ). Consider now the inequality /(* 2 )-/T*i) ^ /(*3)-/(*i) X 2 ~Xi X 3 ~X t This is equivalent to (*3 ~xi)f{x 2 ) - (x 3 ~Xi)f{x{) < (x 2 -x,)/(*3) - (x 2 -x,)/^,) i.e. (x 3 -x 1 )f(x 2 )<(x 3 -x 1 -x 2 +x i )f(x 1 ) + (x 2 -x 1 )f(x 3 ) (9) Solutions to exercises (12.21) 215 which is (9). A similar argument establishes the inequality /(*3) -/(*■) ^ /(*3)-/(*2) x 3 — x, x 3 —x 2 3 LetZG7and Y&J. Set X = f(x) and Y=f(y). From (1) of §12.13, /(ax + ft) < a/(x) + 0/0) = ctX + H Y provided that a > 0, (3 > and a + (3 = 1 . Since /is strictly increasing on /, /"' is strictly increasing on J. Thus ar'(X) + $r\Y) = Qtx+$y<f~ i {aX+m and hence /"' in concave on J. If /is strictly decreasing on /, then/" 1 is strictly decreasing on J. In this case the last step in the proof is replaced by af\X) + PrXY) = ax+Py>r\aX + 0Y) and so/" 1 is convex on /. 4 By the mean value theorem, there exists an t] between x and £ for which /(*)-/«) _ x-% = /'(*?)• Because /is convex, its derivative increases. Thus/'O?) >f (£) '€x > % and/'(r?) </'(£) if* < I Hence Geometrically the result asserts that the graph of the function lies above any tangent drawn to the graph. i-=/U) = /«)+/'«) (-v-S) / concave If/ is concave the inequality should be reversed. The geometry of the Newton- Raphson process is indicated below. 216 Solutions to exercises (12.21) <M*b)<I Given that x„ > %, <t>(x n ) > and <t>'(x„) > 0. (We cannot have <p'(x) = for any x because 0' is increasing and hence we would have <t>'(y) = (y < x)). Thusx n+1 <x n . On the other hand, by the inequality of the first part of the question, *(*n + l) -0(*n) >*'(**)(*,. + l-*n) = -0(*n) and hence (p(x„ + 1 )>0. It follows that £<x n+ , <x„. Since <x„> is decreasing and bounded below, it converges. Say x n -*■ I as n -*■ °°. Then <p(x n ) ■* 0(/) as n -* °° because is continuous and (p'(x n ) -*• <t>'(l +) as n -*■ °° because 0' is increasing. We obtain 1 l *'(/+) and hence 0(/) = 0. Thus I = £. 5 Suppose that /'(I) > 0. Then, using question 4, /(*) > /«) + f'(Mm ~ D - + °° as x ■* + «.. Similarly, if /'(£)< 0, /(*) >/KS) + /'©(•* -£>-* +«»as* -*- ~. It follows that /"(£) = for all values of % and therefore /is con- stant (theorem 1 1.7). 6 Let P(n) be the assertion that / Xi+X 2 + . . .+x n <-{/(*.) + ...+/(*„)} (10) for any x, , x 2 , ■ ■ . , x n in the interval /. That P(2) holds is given. We assume that P(2") holds and seek to establish P(2" + l ). Write m = 2". Then2w = 2" + 1 and ^T^i-4 -(x,+ .. . +x m ) + -(x m + 1 + . . .+x 2m )\ Solutions to exercises (12.21, 13.18) 1 217 1 >■ + ••• +* ■2 f [ * +f/ ■ • • + x, < 2m {fXXl) + " ' + /( * m)} + 2m {f(Xm + t) + ' ' ' + /( * 2 '" )} - ~ {/0f0 + • • • + /(*am)}- 2m We now assume that P(n) holds and seek to deduce that P{n — 1) holds. Write X = x t +x 2 + . . -+* n -l K-l Then / jc 1 +x 2 + ■ ..+X n - i ; \ n-\ = f(X)=f l(n-l)X + X X\ + X 2 + . . . + X n _] + X = / <-{/X*l) + ---+/(*n-l)+/W}. Hence I — -) /(*) <- {/(xi) + /(* 2 ) + ■ • • + /(*„ - 0} and P(n — 1 ) follows. Let a be a rational number satisfying < a < 1 . We may write a = m/n . If & = 1 — a, then = (/; — w)/« . Apply inequality ( 1 0) with x,=x 2 = . . .=x m = xandx m + 1 =x m + 2 = . . . =x n = y. Then /(ax +0>O <«/(*) + 0/00. (11) If a is irrational, consider a sequence <a„> of rational numbers such that a n -*■ a as n -*■ °°. Then n = I — a n -» 1 — a = as n -> °°. We have f(a n x + n y) < a n f(x) + n f(y) (w = 1 , 2, . . .)• Since /is continuous, consideration of the limit shows that (1 1) holds even when a is irrational. • Exercise 13.18 1 If «S H, we have for any x 1 /J = X" n + 1 and hence the result follows from theorem 13.14. 218 Solutions to exercises (13.18, 13.26) If n is replaced by a rational number »#— 1, the same argument applies. The restriction < a < b is required because x a need not be defined for* <0 (e.g. if a = — 2,;c 0: is not defined for* = 0). If a = — 1, the argument fails. (Why?) 2 The integrand is not defined when x = 1 and, in fact, has an unpleasant 'singularity' at this point. 3 As in example 13.17, we obtain ^ n {0 a +l Q + ... + («-lf} <j 1 o x«dx<-^- 1 {\" + 2 a + . .. + ««}. It follows that and the result follows. • Exercise 13.26 1 Suppose that, for some % G [a, b],g(%) > 0. Then, because # is con- tinuous on [a, b] . a subinterval [c, d\ containing £ can be found for which f(*)>k(D (c<x<d) (proof?). But then j a g(t) dt > j g g(t) dt > \g(£)(d - c) > 0. 2 We integrate by parts. Then J b a xf"(x)dx = [xf(pe))*-f*f'(x)dx and the result follows. 3 Observe that F'(t)=f(t) > {F(t)} xn (theorem 13.12). Hence (x - 1) = f 1 dt < f {F(t)}' U2 F\t) dt (theorem 13.23) Solutions to exercises (13.26) 219 = {2{F{t)Y n ] x i (theorem 13.14) = 2{F(x)y 2 <2f(x). 4 By the continuity property (theorem 9.12), the function /attains a maximum M and a minimum m on the set [a, b]. Hence, by theorem 13.23, I j^ g(t) dt < J o " f(t)g(t) dt<M j a " g(t) dt. (12) We now appeal to the intermediate value theorem (corollary 9.10). The function F defined on [a, b] by F(x) = /(*) f g(t)dt Ja is continuous on [a, b] and hence a % S [a, b] can be found such that F{%) = f t\t)g(f)dt J a because of (12). Ug(t) - 1 (a < t < b) and Fis a primitive for /on [a, b], then the result reduces to F(b)-F(a) = F'(£)(b-a) which is the mean value theorem (i.e. theorem 1 1.6) for the function F. 5 We have j a b f(t) g (t) dt = [f(0G(t)) b a - jl noG(t) dt. Since f\t) > (a < / < b) we may apply the previous question and deduce the existence of | E [a, b] such that j" a t\t)g{t)dt = [/(/)G(f)L 6 -G«)J" o V(0 df = [f(0G(t)] b a -G(k)im] b a = /(«) Jj«05 ^ + /(*) j { %« *'• 6 We integrate the given integral by parts. Then 220 Solutions to exercises (13.26, 13.34) ■ _ t \n-lf(n-W f - i ,x [(x-ty-v^Xm + J* (n-\)(x-t)"- 2 f ( -"- 1) (t)dt 1 («-!)! (* - sr ~ l r~ l w -...-(x- a/'® + f ,. But ^ = \l f\t)dt =m-rm and therefore /(*) «/(© + (« -!)/'(!) + ... 1 as required. Observe that, by question 4 above, fc-8f , ~ 1 /*-*to+J$, i /^to) (7 (x-tf^dt = ^r^/ (n) (r?) (h-1)! V, 'J* R! for some value of tj between x and £. • Exercise 13.34 1 (0 J„ "' J X .,, = f-2(!-x)" 2 ]'- 5 =2-25" 2 ^2as5->0+. «/. g o V(l -x) * x 2 dx 3\-llX (i + x y = [-ki+*r 1 io 111 1 — — -»■ — as X •* + **, 3 3(1+ A -3 ) 3 (iii) Since (1 4-x+;c 3 )(l +x 7 ) \ +x+x s we can find an H such that 1 as x •* + °° Solutions to exercises (13.34) 221 1 + x + x 3 < H l+x+x s \+x- (x>0). The existence of the improper integral then follows from proposition 13.29 (see example 13.30). (iv) £"V5f = **-**"*** = .4- 2 ^ +ooas5 - o+ - (v)Put 1 -x=y.Then r i-8 dx f & — dy _ r 1 dy_ y J 6 y V§ j-i-6 dx i-6 — dy _ r 1 Jo 1 -y ~ Ji v Js 1 -x and so (v) reduces to (vi). (vi)Put.y = 1/z. Then i-i dy r 1 zdz _ f 1/5 dz J& ~y~ = Jl'6 ~~^~ "Jl Z by example 13.33. -°°as8->-0 + 2 / = i: '•2 + >- 2v — 1 r 1/2 2*- dx = 1/2 -y x(\ -x) f 1/2 2x-i r 1 Ji/2-y v(i -x) J" 1/2+ y 2x-i x(l -x) dx. Put .v = k - « i" the first integral and jc = \ + u in the second integral. Then ' - -J. ^kTht)"" + Jo «r^r^) d " = °- Ho\ r However, the improper integral does not exist. We have -1/2 2sr— 1 f2s> ucto '•-^f i/2-y x{\ —x) (1-Z))(1+U) = _r Jo 1 1 dv l—v 1 + v\ o°asy^-\— (exercise 13. 34(1 v)). On the other hand -1/2+2 2x — 1 r J 1/2 jc(1 -x) dx = +2 j 2y urfzi (1-0)0 +v) - + °° as z ■* t — . 3 After theorem 13.32 it is enough to establish the existence of the im- proper integral °dx_ x a s: 222 Solutions to exercises (13.34, 14.3) But r*dx x rxfa _ 1 X a-1 X"- 1 a- 1 asZ- 1 -a provided thai a > 1 . • Exercise 14.3 1 We have log 2 = {* ^>i(2-l)>0. Observe that log (2") = n log 2 ■* + °° as n -* °° and hence the logarithm is unbounded above on (0, °°). Since it is strictly increasing, it follows that log* -* + °° as x -*■ + °°. Also log (2"") = — n log 2 ■* — °° as « -» °°. Thus log a: ■*■— °°asx->0+. 2 Since the logarithm is concave, logj>-logl<(j/- 1) for all y >0 (see exercise 12.21(4)). Take y =x s . Then, using theorem 14.2(a), s log x = log (x s ) <x s — Kx s . (i) Given r > 0, put s = \r in the above inequality. Then x~ r \o%x = x- r ' 2 (x- s \ogx)<x- rn /s--0asx-+ + c°. (ii)Puty=x~ l m(i). 3 We have F'(x) = log x + x . x~ { — 1 = log x and hence F is a primitive for the logarithm. By theorem 13.14, | o ]og(\+x)dx =J i log /</? = [t log t-t]t = 2 log 2 - 1 = log (4/e). As in example 13.17, £M ,+e ?H*< Hence I + x) dx < T - lo = i n k = i k n Solutions to exercises (14.3) 223 > + BH" n- \ n l/n ^ <log {(l+- n log (w + 1 )(« + 2) ... (2« — 1) I," '<7 <log fa + 1) . . . (2k) l/n (w(2/i)! log., 2n.n\ It follows that l/n -<tog - <l0| '(2«)! h! l/n 0<log- (2»)! 4\ 1 — log — < — log 2 ■* as n -*■ °°. 4 Consider the function 0: (0, °°) -*B8 defined by <p(x) = log* —ax. If a < 0, </>(x) -» —oo as x -* + and (p(x) -* + °° as x -* + °° (exercise 14.3(2)). It then follows from the continuity of that a solution of the equation $(x) = exists. The case a > is somewhat more interesting. It remains true that <p(x) ^ —°° as x -* + but, for this case, 0(.*) ->— °°asjc -* +°° (exercise 14.3(2)). Consider the derivative <p'(x) = --a. From this identity it follows that is strictly increasing on (0, I/a] and strictly decreasing on [1/a, °°). Thus *(*><* (| -tog~l ° 3) witli equality if and only ifx = 1/a. Because <j> is continuous on (0, •*>), solutions of <p(x) = exist if and only if the maximum value 0(1 /a) of 0is non-negative, i.e. log(l/a)>l = loge. ( 14 ) Since the logarithm is strictly increasing, (14) holds if and only if l/a>e,i.e. ea< 1. The inequality e \ogx<x(x> 0) is simply (13) with a = 1/e. We prove that the sequence (x n ) is decreasing and bounded below by 224 Solutions to exercises ( 14. 3) e. We have X„+i = elogx„ <x„ and so <x„> decreases. If x n > e, then jc„ + i = elogx„>eloge = e and so the fact that <x„> is bounded below by e follows by induction. Suppose that x n -*■ f as n -*°°. Since x n+1 = e log x n and the logarithm is continuous, £ = elog£. But tliis equation holds if and only if % - e. 5 Apply theorem 13.32 with/(x) = 1/x. Then 1.1. .1 r n dt A„=l + i + i + ... + i-f^ 2 3 n Ji { and A„ -> y as /? -* °°. As in the solution to exercise 6.26(6), 111 11 1 + +... + - = — 234 2n - 1 2n n+\ n + 2 1 1 + ... + — 2« 2 2n 2 n = {log 2n + A 2n }-{log« + A„} = log2+A 2n -A n ■* log 2 + y — 7asn-*°°. 6 {i)D{\ogx} s =s{\ogx} s - l \lx (ii)0{loglogx} = If r 96 1, take x - 1 = -r in (i). Then 1 1 j2 r(logx)'" dx = 1 -r (log*) 1 x i(log 2) 1 r-1 logx X (r<l) is A" ■* + °°. If r - 1 , we use (ii). Then * 1 I. x log* dx = [log log xj 2 '"* + °°asx ■* + °°. The appropriate conclusions about the given series follow from theorem 13.32. Solutions to exercises (14.5) 225 • Exercise 14.5 1 (i) Put X = exp x and Y = exp y. By theorem 14.2(f), log AT = log X + log Y = x+y. Hence exp(x+j<) = exp {log XY} = XY = (expx)(exp.y). Alternatively, consider D ex<p(x+y) exp x exp (x + y).expx — expx.exp (x + v) (exp x) 2 = for a fixed value of v. We obtain exp (x+y) expx = c (xGR) for some constant c. The choice x = shows that c = expy. (ii) Put X = exp x in theorem 14.2(H). logA" r = Hog A = rlog (expx) = rx. Hence exp(rx) = exp (log X r ) = X r = (expx) r . 2 (i) By exercise 14.3(21), for any s > 0, A~ 5 log A^0asA^ + °°. Hence (log AT r A = {(log A) A~"T r ^ + °° as A ^ + <=•=. It only remains to put X = exp x (see exercise 8.20(2)). (ii) By exercise 14.3(2ii), for any s > 0, A s log A ■* as X -* +. (log A)'"A = {(log X)X 1,r } r ■* as A -> +. Put X = exp x again and the result follows. expx — 1 , expx 3 (i) lim — = hm —f- = expO = 1. x — X x-*0 1 ,.,, .. log(l+*) ,. (1+*)'' _ 1 (11) hm = hm ; = — — - w * -► x * -* 1 1+0 = 1. 4 We have 226 Solutions lo exercises (14.5) H\t) = li(t)>H(t). Hence T* H'(t) c x Jo m dt> >* Udt = x (x>oy Thus [logH(t)] x >x. But H(0) = 1 and it follows that log//(x)>x i.e. h(x) > H(x) > exp x. 5 We have /'(*) = 18.3a; 2 exp (1 8x 3 ) f"(x) = 18.6xexp(18x 3 ) + (18.3) 2 x 4 exp(iax 3 ). The sign of f"(x) is the same as that of 4>(x) = 2x + 18.3x 4 = 2x{\ + 27x 3 ). Observe that 0(0) = <t>{—\) = 0. Since 4>'(x) = 2 + 8.27x 3 = 2 + (6x) 3 , it follows that <p is strictly increasing for 6x > — 2 1 ' 3 and strictly decreasing for 6x < — 2 1 ' 3 . Thus is non-negative on the intervals (— °°, — |] and [0, °°) and so /is convex on these intervals (theorem 12.19). But is non-positive on [—3, 0] and so/is concave on this interval. 6 Since /increases on [0, °°) f(x)<f(k) (k-l<x<k) f(k)<J\x) (k<x<k+l). Hence £, f(x)dx<f(k)< I* 1 max. But J mdx = £ J ft _ i f(x)dx; J, /(x)dx = £ J ft /(*)<&. ft — 1 k — 1 Apply (he result with/(x) = log x. Solutions to exercises (14.5, 14. 7) [x\ogx-x\ n < £ \ogk<[x\ogx-x\r 1 « log n-n< log «!<(« + 1 ) log (rt + l)-(« + 1)+ 1 nlogn-log«! <«<(«+ l)log(n+ 1) — log/j! 227 (15) log R! — <n<log (« + ir n (« + i)" +1 — < exp /? ^ • «! n (Note that (1 5) involves the use of an improper integral on the left hand side. From exercise 14.3(2), x logx -* asx ■* +.) • Exercise 14.7 1 (i)c?* y m exp {(x+y) logo} = exp {x log a + y log a} = exp (x log a) exp (y log a) (exercise 14.5(H)) (ii) (abf = exp {x log {ah)} = exp {x log a + x log ft} (theorem 14.2(i)) = exp (x log a) exp (x log b) (exercise 14.5(H)) = a x b x . (m)a~ x = exp {-x log a} = (exercise 1 4.5(1 ii)) exp (x log a) J_ ~ a x ' (iv) Note that, since a x = exp (x log a), it follows that X log a = log (a*). Hence a xy = exp {xy log a} = exp {y log (a*)} = (a x ) y . 2 We have Z) {exp (x log a)} = log a . exp (x log a) = (logff)a*. 3 By theorem 8.9 and exercise 14.5(3ii), 228 Solutions to exercises (14. 7) n log(l + xn' 1 ) = x JogO+xn- 1 ) xn ■ x as n -* °°. Since the exponential function is continuous at every point, it follows that 1 +-) = exp « log 1 +- exp x as n 4 By exercise 14.3(2i), n l log n •* as n -* °°. Hence, log n) •* exp = 1 as n -» °°. n v " = exp 5 (i) Consider j^~°e' xl/2 dx. Take <j)(x) = e~ xn in proposition 13.29. We know that ^ e~ xn dx = [-2e~* n ] x ->2asX-*°° arid A similar argument (with tp(x) =e x ' 2 ) establishes the existence of r' e-^dx. J-*- DO It follows that the improper integral L— * * exists. (ii)See §17.3. (iii) We have Jo ] + AT : dx = r£iog(i+x 2 )js 2\1.Y = | log (1 + A" 2 ) ^+°° as *->•+«>. Hence the improper integral xdx + .v 2 Solutions to exercises (14.7) does not exist, even though 229 /. -x 1 +*' = 0->-0asA'^+°°. 6 (i) The trick is to differentiate the given equation with respect to x keepings constant, and then with respect toy keeping* constant. For each x and y, we obtain f'(x) = f'(x+y) =/'<» and it follows that f\x) = c (xGR) for some constant c and hence that f(x) = ex + d. Since /(0) = 2/(0), we must have d = 0. (ii) As in (i) we obtain /V)/00 - /0)/'0) and therefore fix) = c (xGR). Hence log/(x) = G* + rf. Since /(0) =/(0) 2 , d = and hence fix) = e cx (x S R). (iii) As in (i) we obtain /'(*) _f(y) y x and therefore f\x)= C - (*>0). Hence fix) = clog* +d (x>0). Since/(l) = 2/(l),d = 0. (iv) As in (i) we obtain 230 Solu tions to exercises (14.7, 15.6) /'(*)/(?) _ f'(y)f(x) y x and therefore /(*) x log/(jc) = clogjc + d. Since /(l) = f(\f,d = and hence f(x)=x c (x>0). • Exercise 15.6 1 (i) The interval of convergence is R because |<7 n |"" = — >-0 as«^-°°. n (ii) The interval of convergence is {0} because (n + 1)! = (« + 1 ) ** + °° as n (iii) The interval of convergence is U. This is most easily seen by com- paring the series with the Taylor series expansion for exp x. (iv) As (iii). (v) The interval of convergence is (—4, 4). The radius of convergence is equal to 4 because (n+1) 2 (« + l) 1 ■ — as n -*■ °°. 4 (2« + 2)(2#i+l) 2(2« + l) Note that the series cannot converge at the endpoints x = 4 and x = —4 of the interval of convergence because <<z„4"> is strictly increas- ing and hence cannot tend to zero. We have » — = r > 1 (w = 1 1 2, . . .). a„4" (« + J) (vi) The interval of convergence is [— 1, 1). The radius of convergence is equal to 1 because v« V(» + 1) ■* ] as n -*■ °°. The series does not converge for x = 1 . The series converges for* = — 1 by theorem 6.13. Solutions to exercises (15.6) 231 2 To employ exercise 13.26(6), we need the nth derivative of the ftlBO tion/:(-l,°°)-»-R defined by /(*) = log (1 + x). We have fix) = 1 +x f\x) = - (i+*r f<"\x) = (n-l)\ (-1) n-1 (1+x)"- (A formal proof would require an induction argument.) These deriva- tives are evaluated at x = and from exercise 13.26(6) g(i+^) = /(o) + f [ /'(0) + ... + ^ i / ( "- 1) (o)4-^ lo where = x + - — ■ • • + (-1)" (H-l) + En E » - ==niir *-« , -w« (8-1) 1 dt — 2— re (n-l)! Jo v x-ff^H-l)! (-1)"" 1 dr. (1 + 0" These formulae hold for any x > - 1 . We are concerned with the values of x for which E n ->• as « -> <*>. We therefore have to consider I4J = Jo (1+0 ffO<x<l. ^=i; < ( 1+ 'V < " < i'» ( '-' r '* x" „ = ► as m -* °°. n If-Kx<0, f o (r -*)""' It is ea: ily seen that 232 Solutions to exercises (15.6 j .t—x 0<- -<-x (-Kx<t<0) 1 +t and it follows that, for - 1 < x < 0, \E n \A x (-xf * — = (— ^)" _1 [— log (1 + *)]-»■ as w ->• °°. The series expansion for log 2 is now obtained simply by setting.* = 1. 3 To obtain the form of the remainder as given in theorem 1 1 .10, we have to write x = 7? in the formula for/ {n) (x) in the previous question. This yields E„ = (-D n-l 1+TJ where r\ lies between and x. If < x < 1 , there is no problem. Since 1 + 7J>1, n If — 1 < j: < 0, we can assert no more than But —x 1+x R \1 + X >1 (-1<jc<-|) and hence we are only able to show that the remainder term tends to zero for values of x satisfying — ^ < x < 1 . 4 We have Rn = 1 1 (n + 1)! (« + 2)! 1 1 + 1 ■ + • 1 < (n + 1)! 1 (n + 2) (n + 2)(« + 3) 1 + . . . (« + 1)! 1 (« + 1)! (n +2) (n + 2) 2 1 + .. . !-(« + 2) i-i Solutions to exercises (15.6) 1 n + 2 25.? (« + !)! n + 1" From the power series expansion for e x with X = 1 , we know that 1 1 1 0<e-l+— + — + ...+ — 1! 2! H! = R„. We have to take n = 4. Observe that 16 16 1 R t < 5! 5 120 5 100 5 Suppose that e = m/n where m and n are natural numbers. Then 1! 2! Rj is a natural number. But, by the previous question, n! (n + 2) 0<7 = n\R„< (n + l)!(n + 1) ( W + 2 ).<1 (n + l) 2 provided that n is a natural number. But there are no natural numbers in the interval (0, §) and hence we have a contradiction. (Note that n + 2 1 ■ + ■ 1 (n + l) 2 (r + l) 2 ' (n + 1) and hence is obviously decreasing.) 6 We prove by induction that, for any x 4* 0, f«Xx-) = P n ^\e-^ (16) where P„ is a polynomial of degree 3n. For n = 0, there is nothing to prove. Given that (16) holds, / (n + °(*) = K -^K^ + Mt)-^ = ^HH& : Observe that the polynomial P n + 1 defined by P n+1 Q>) = -y 2 P' n (y) + 2y 3 P n (y) has degree 3n + 3 provided that P„ has degree 3n. This com- pletes the induction argument. 234 Solutions to exercises (15.6, 15.10) We next consider the existence of the derivatives / <n> (0). Again we proceed by induction. We show that/ <n) (0) = (n = 0, 1, 2, . . .). This is given for n = 0. On the assumption that/ ( " _1> (0) exists and is zero, /<"-»(*)-/t"-'>(0) i M _ v< x x ""' \x It follows that lim x -» lim j>/ViO)e~ 5 ' = by exercise 14.5(2) (exponentials drown powers). The same argument also applies with the left hand limit. It follows that/ (n) (0) exists and is zero as required. Since /„(0) = (n = 0, 1, 2, . . .), the Taylor series expansion of/ about the point 0, is + + + + + .. . This converges tof(x) only when x - 0. • Exercise 15.10 1 The radius of convergence R of the power series fix) = I «("-l)--(«-"+l) JC „ n=0 is given by n! rt! ^ „™. (« + l)! |a(«-l)...(a-/i + l)| .. |a-n| = lim = 1. «-»■« « + 1 For |jc| < 1 we may therefore apply proposition 15.8 and obtain = «a-l) »-» + !) Hence, (1 +*)/'(*) - a(tt-l)...(a-w+l) ~ tt(a-l)...(tt-H+l) „ A (»-D! 4l («-D! " I a(a— 1) ... (a— w) a(a — 1) . .. (a — n+ 1) = a + ) ( 1 Solutions to exercises (15.10) a(a — 1) . . . (a — n) 235 = «+ I («-!)! U-L n a—n = a i+ - tt(tt-l)...(a-/i +1) ^,1 «! It follows that, for \x\ < 1, D{(\ +xy a f(x)} = -a(I +x)- a -'/(x) + (l +x)- Q /'W = and therefore f(x) = c(\+xf (|x|<l) for some constant c. But /(0) = 1 and hence c = 1 . 2 The radius of convergence R is given by 1 r (n + l) 2 , — = hm ; — = 1. R n - - n 2 We know that n = Hence, usi oo 1 — X (W<1). Hence, using proposition 15.8, 1 (1 -^) 2 CM<1) 2 X «(«-l)x"- 2 = — — - 3 (|x|<l). Therefore, £ «V = J {n(n-l) + n}x n n=l = X 2 . £ «(«-l)x"- 2 + X. J BX"" 1 2x 2 n = n = 2 .2 + (1-x) 3 (1-x) 2 2x 2 +x(l + s) _ x(l+x) (1-x) 3 = (1-x) 3 (W<1). 3 The power series 236 Solutions to exercises (15.10) _y + i has the same radius of convergence as the given power series because lim sup lim Lap \a n \ (« + I) By proposition 15.8, if x e / but is not an endpoint, F'(x) = f n% (« + 1) .(n+l)x n = f{x). Hence fo-f(o) -£/&#*. Observe that F(0) = 0. 4 The function/: (— °°, l)->-|R defined by log(l-*) fix) = -- X has the Taylor series expansion /(*) = 1+I + J + J+--- (-1<*<1). (Note that formula (1 7) does not make sense when x = 0, but it is natural to define /(0) = 1.) If < y < 1 we obtain from the previous question \ f(x)dx=y+f 2 +f 3 +... and hence (17) r 1 s, v" lim /{*)<& = lim £ V But the final power series converges for .y = 1 , and from proposition 15.8 we know that the sum of a power series is continuous on its inter- val of convergence. Thus -/. ' log(l-*) 1 x „=i n 5 We consider the power series *(*)= I («»-*„)(* -8". n=0 Solutions to exercises (15.10, 16.3) Ifgix) = (x € /), then, using proposition 15.8, = g(£) = a -b = g'm = l(«i-*i) = /(£) = 2(a 2 -b 2 ) = *'"«) = 3.2(a 3 -6 3 ) 257 and the result follows. 6 By proposition 15.8, n = From the previous question it follows that a n _, = «a n («= 1,2, .. .) as required. But then 1 1 22 = ^1 = — «0 and, in general, a „ = ^ (« = 1,2,...). n\ Thus /(*) = "o I -T = «o**. n=0"l • Exercise 16.3 2 4 1 (i)cos0= 1--+--...-1 3 5 (ii)sinO = 0- — + — -. ..= (,-xf (~xf (iii)cos(-x) = i--^r + -^r — . . . = cos X 238 Solutions to exercises (16.3) (iv)sin(-*) = (-*)- (-*) 3 , (-*) 5 3! 5! . . . = — sin x. We appeal to proposition 15.8. Then Dcosx = D 1 + 2! 4! _ 2* 4* 3 6x s 2! 4! 6! Dsin* = D{x — — + — — , = -x+ 3! 5! . . . = — sin x. 3! 5! = '"i + Ij----" 5608 *- 2 We have £'(*) = cos (x+y) — cos x cos y + sin * sin y = ft (*) «'(*) = — sin (x +7) + sin* cos j> 4- cos* sin 7 = —g(x). Hence D{g(xf + h{xf) = 2g(x)g\x) + 2h{x)h\x) = 7g(x)h(x)-2h(x)g(x) = for all x. It follows that g(xf + h(xf = g(0) 2 + h(0) 2 = for all x. Since {#(*)} 2 > and (M*)} 2 > 0, we may conclude that, for all values of x, m = and «(*) = 0. 3 (i) 1 = cos(*-*) = cos*.cos(-*)-sin*. sin(-x). (iii) < cos 2 * < cos 2 * + sin 2 * = 1 . Hence |cos x\ < 1 . Similarly |sin x| < 1. (iii) sin 2x = sin x. cos x + cos x. sin x (iv)cos2* = cos x. cos* —sin*. sin*. . ... .. sin* cos* 4 (1) hm = lira = cosO = 1. x -* * X -► 1 Solutions to exercises (16.3, 16.5) 239 1 — cos * sin * 1 (ii) lim ; = lim — - = — . x-*o x *-»o 2* 2 5 By the mean value theorem, there exists £ between and * such that sin * — sin *-0 = cos £ and we know, from question 3 above, that |cos £| < 1. The convergence of E~ =1 sin (l/« 2 ) may be deduced from the com- parison test because sin \n <4: («=1,2,...). Alternatively, we may appeal to exercise 6.26(2). From Question 4(i) above lim sin(l/« 2 ) = l; sin(l/«) lim : = 1. I//1 2 "*-*"" 1/n The latter result shows that S~ = , sin (1/n) diverges. 6 From exercise 13.26(5) it follows that, for some % between m and n, r" sin r 1 rJ 1 r " r" sinr 1 r( , , 1 f" . dt = — sin t dt + — sin t Jm [ m Jm n Jf dt = — (cos m — cos ?} + — {cos if — cos n\. m n Hence r" sin t Jm ( dt < m Let e > be given and choose N = 4/e. Then for any n>m>N, r" sin / f" sin / r m sin t , dt-\ Ji t Ji f dt 4 4 and it follows that we are dealing with a Cauchy sequence. Hence f sin t lim „-»°o Ji ? exists (theorem 5.19). • Exercise 16.5 1 From the discussion of § 1 6.4 we know that cos * > (— \ti < * < \n ). Since sin (* + jn) = cos*, it follows immediately that 240 Solutions to exercises (16.5) sinx>0(0<x<7r). But D(cosx) = — sin* and thus the cosine function decreases on [0, »r]. Because cos (x + ir) = — cos x, we may also deduce that the cosine function increases on [ir, 2ir] . Further, D 2 (cosx) = — cosx. Thus Z> 2 (cos x) < (— %n < x < \it) and so the cosine function is con- cave on [ _ |ff,|jr]. Because cos (x + n) = — cos x, D 2 (cos x) > (jn < x < §7r) and so the cosine function is convex on [\it, §jt] . 2 We have tan (x + n) = sin (x + it) sin x cos (x + n) — cos x (provided that cos x ^ 0). Also cos x. cos x — sin x (— sin x) = tanx .Dtanx = cos 2 x cos 2 * (provided that cos x ¥= 0). Since cos x does not vanish on (— \n, |jr), it follows that tan x is strictly increasing on (— § jt, ^it). We have cos x -► as x •* \it — and sin x •* 1 as x -> ^w — . Since the sine and cosine func- tions are positive on (0, |ff), it follows that sin x tanx = »-+ 00 asx-+57r— . cosx Similarly for tan x -*■ — °° as* ■* — ^rr +. 3 We have £>sinx = cosx>0 (— |»<x<Jff) and hence the sine function is strictly increasing on [ — 5^, \n\. Since sin (— \t() = — sin (£*•) and sin (\it) = 1 , it follows that the image of .v = arcsin y Solutions to exericses (16.5) 1 1 241 D (arcsin v) = , D sm x cos x where x = arcsin .y, i.e. sin x = y. Since cos 2 x + sin 2 x = 1 , we have cosx = ±V(1 -sin 2 x) = ±V(1 -y 2 )- We take the positive sign because we know that the arcsine function increases. Thus D (arcsin y) = 1 (-Ky<\). V(i -y 2 ) We have Dcosx = -sinx<0 (0<x<tt). Also cos (0) = 1 and cos (ir) = — l. Hence the image of[0, n] under the cosine function is [— 1 , 1 ] . £>(arccosj>) = I 1 -1 Dcosx — sinx \/(l ~y 2 ) We take the negative sign because the arccosine function decreases. Note that cos (x + ^tt) = — sinx = sin(— x). If x lies in the range [— \-n, \ti] we may write — x = arcsin^. But then cos (x + ^7r) = ;'. Since x +\n lies in the range [0, ir) it follows that x + %ir = arccos>>. i.e. \u = arccos>> + arcsin y. 4 It follows from question 2 above that the tangent function is strictly 242 Solutions to exercises (16.5) increasing on (— \n, \-n) and tan x -*• + °° as x •* \n — and tan * as x-*— \tj+. Thus the image of (— \ti, \ti) is R. 3' = lan .v jrX = arctan v -w ";■ Z) arctan J» = 1 = cos 2 * (y = tan *). D tan x But cos 2 x + sin 2 * = 1 and so 1 + tan 2 * = 1/cos 2 *. Thus 1 1 D arctan y = 1+tan 2 * " 1 + y 2 ' Finally, C x dx r- + * 2 = [arctan *] = arctan X-*- — as X-* + °°. 5 Differentiate with respect to* keeping y constant and then with res- pect toy keeping* constant. If*y < 1, f {x) m f ( x+ y\ f ] (i -*y)-(x+y)(-y) (\-xyf Hence f'(x) - /' Also , (x+y 1 —xy 1 +y 2 x +y (1 -xy)- 1 +* 2 ,2 ' '<» = '" \i -*,/ id -xy?\ ■ Thus /'(*)(! +* 2 ) =f'(y){\ +y 2 ) (xy < 1). Hence fix) = 1+* 2 (Xi Therefore /(*) = c arctan * + d (* E M). But 2/(0) = /(0) and so d = 0. Solutions to exercises (16.5) 243 If* = y, we obtain »-£'(&) (J<>1) ' We apply this formula with/(*) = arctan *. Since arctan y -* j7r as y-> + °°> ( 1 ( 1* arctan 1 = lim {arctan*} = hm —arctan I- _ j = U- The power series * — x 3 x 5 — + . . . 5 5 is easily seen to have interval of convergence (— 1 , 1 ] (for * = 1 , use theorem 6.13). We apply proposition 15.8. Then, for |*| < 1, D * 3 X 5 arctan *—* + — — — +... i | v 2 „4 | - ]+x2 11**1 ... i i 1+* 2 1+* 2 It follows that * 3 * s arctan * = C + x— — + — — . (W<1). But arctan = and so C = 0. We know from proposition 15.8 that the sum of a power series is continuous on its interval of convergence. Hence \-n = arctan 1 = lim_ {arctan*} x 3 x s = lim *- T + --... * -» t - 3 5 — I 3 + 5 6 (i) Consider the sequence <*„> defined by * n = {i7r + 2«7r}-'. Then *„ -*■ as n -* °°. But /(*„) = sin(|ir + 2n7r) = 1 -/►/(()) as «• 244 Solutions to exercises (16.5, 1 7.4) (ii) We have \g(x)\ < |x| and hence g(x) -* as x -> by the sandwich theorem. However, g(0+x n )-g(0) *•> g(0-x n )-g(Q) •* 1 as n •*• °° —x„ -*■ — 1 as n •* °°. (iii) We have \h(x)\ < x 2 . Hence h(x)-h(p) -0 < Ixj ->0 as;t->-fJ and hence m'(0) = by the sandwich theorem. • Exercise 17.4 1 We have < sin x < 1 (0 < x < ±tt). Hence 0<sin n + 1 x<sin"x and therefore 0</ n+1 </ n (n = 0,1,2,...). To obtain the recurrence relation nl n = (h — l)/„- 2 , we integrate by parts. For n > 2, /„ = J o sin"* = sin x. sin" 'jtdx = [-cosx.sin"- 1 x]S /2 +n cosx.(n-\)sin n - 2 x.cosxdx Solutions to exercises (1 7.4) 245 - (n - 1) J"' (1 - sin 2 *) sin"- 2 * <£c = (n-1 ){/„-»-/„}. i.e. n/„ = («-l)/„-a (« = 2,3,...). It follows that hn Jin-i 2m + 1 l<-^< ■-*■ 1 as n -> °°. An + l ^2n + l 2« 2 Use (18) of the previous question. We obtain r (2m ~ 1) . (2h-1)(2w-3) t /*„- 2/? li»-a- 2 „ (2w _ 2 ) i2 '- 4 _ (2«-l)(2n-3)...l 2k(2m-2)...2 ° _ 2m(2m - l)(2w - 2)(2« - 3) ... 2.1 r*« (2m) 2 (2h-2) 2 ...2 2 '• lo -J^f c»-o,,. 2> ...,. 2m 2«(2n-2) 2« + l '"' (2m+1)(2m-1) 2h(2w-2)...2 (2m + 1)(2m - 1) ... 3 ' (2m) 2 (2m - 2) 2 . . . 2 2 - »« / 2 n-3 •f Jo (2m + 1)(2m)(2k-1)...3.2 (2 B «0 2 r ..,„/. (2 n «!) 2 (18) [-cosx]5 /2 = (2m + 1)! sin x <& (k=0,1,2,...). (2n + 1)! It follows that l 2n _ (2k)! n (2n+l)\ -n {(2m)!} 2 /a„ + i (2"«!) 2 " 2 " (2"m!) 2 2 v ~" * y {2"m!}' in i n {co) 2n+i/ y 2 "} 2 ~2 ( J {2 n Cn" +1/ V n } 4 ,2 4 "2.M 4 " + 1 e- 4 " 7T = ^(2"+OC" a " 2 4n"' B 4n^ e ^n 2m+ 1 n C 2 -*-2nC~ i asn^-°°. By the previous question, 27rC 2 = 1, i.e. C = y/(2ir). 246 Solutions to exercises (1 7.4) 3 The fact that (d n — (12m) '/increases was demonstrated in §17.2. As in §17.2, x 1 x 4 x 6 I 1 *-*♦,-/» -j+ 7 + 7 +... X- — r Hence 1 x* 1 d n -d n + l >- = 3 (2w + 1)2 But 1 12m 2 + 12m + 3 1 1 12m + 13- 12m -1 12m + 1 12(/i + 1)+1 (12m + 1)(12« + 13) _ 12 " (12m + 1)(12m+ 13) ' It follows that i d n ~ 1 12m + 1 - *,♦»- l 12(« + 1)4-1 > (12m + 1)(12m+ 13) - 12(12m 2 + 12m + 3) 3(12m+1)(12m+13)(2m + 1) 2 The numerator is equal to 12m(14-12) + (13-36) = 24«-23>0 (m = 1,2, ...) and therefore the sequence (d„ - (12m + l)""') decreases. We know that d n ■+ d as n -> °° and therefore (m = 1,2,...). d n ~ -T < d < d n 12m " 12m + 1 Taking exponentials and noting that C = exp d = s/(2ir), we obtain '<- ^g-l/Oln + l) n< as required. We may deduce that 1(1 _„-l/(12n + D m!(1 )< m! - y/(27T)n n + U2 e-" < m!(1 -e" 1 ' 12 "). Since e x — 1 is approximately equal to x for small values of x (Why?), the maximum possible error in approximating to n! is about 12m n\. Solutions to exercises (1 7.4) 247 When m = 100, this is small compared with m! (though, of course, it will be a very large number compared with those usually encountered). 4 Consider, for < 5 < A, the value of j s f->e-<dt = — e x -t + r. - e' 1 dt = -{A x e- A -8 x e- s } + - C fe'* dt. x x J & Observe that A x e' A -* as A -* +°° (exponentials drown powers) and 5 x e~ s -* as 5 -* +. It follows that rw = -r(x + l) as required. 5 Let0<a<a<jc<^<6</3andlet0<5< A. Then {!>->'**< f y-i „-t e' c dt <J" S I**" 1 -/*"*!*"**. We may apply the mean value theorem and obtain r*-i_,y-i x-y = Qogtytt-* (19) (20) for some £ between x andy. For any r > 0, t~ r log f -*■ as t ■* + °° and r r log r ■* as r ■* +. It follows from (20) that we can find an H such that \t x - 1 -t y - 1 \<H{t°- 1 + t p - x }\x-y\ and hence it follows from (19) that TO - r(y)i < h\x -j>i{r(oo + tm) and the continuity of the gamma function then follows from the sand- wich theorem. 6 We show that log V{\x + \y) < \ log P(x) + \ log T(y) and appeal to exercise 12.21(6). By the Schwarz inequality (theorem 13.25), if 0<5<A, j" 5 A ,U* y -2),2 e - t J m | 6 A (^-0/2 e - ( /2 )( ^-l)/2 e -(/2 )A ) < t*-\ „-t e~' dt St"-"- dt 248 Solutions to exercises (1 7.4, 1 7. 7) Thus and the result follows. • Exercise 17.7 1 Make the change of variable — t = log u in the integral .A j f^e-'dt. 2 We use the inequality + nfn\ = Cv + n) y r(n + 1) < TO + « + 1) < (" + 1)T(« + 1) ■ (n + l) y w! (21) established in the proof of theorem 17.6 for < y < 1 and « € M. Given x, we let n + 1 be the largest natural number satisfying n + 1 < x and write x = y + n + 1 . Then < y < 1 . Consider r(*) < (« + l) y «! (« + l)V(2w)n W" y/(2Tt)x x x- U2 e- x y/(2ir)x"x "' X x- U2 e- X y/(2n)x x x- in e- (/! + l)WVV" (y+n+\) y+ "(y+n+iy ,2 e l/2„-y-n-l ■* 1.1. jy + « + 1 1 n 1 + y + i ,y + i >> + « + lj e y * ' = 1 as n -*■ °°. ,y + i A similar argument for the left hand side of inequality (21) com- pletes the proof. 3 We have (log(l +z)-z lim *-* o Consider = lim z -► = lim Z-+0 d+z)-'-i] 2z -(1+z) ■i) ^ Solutions to exercises (1 7. 7) y/u f u (U + Xy/u) - ^ (« + Xy/uf^ <T" "* V " y/u{u+Xy/uy- 1 e- u -*^ u V(27r)« u w- 1/2 e- u \u-l 249 V(2tt) " I y/u 1 -l »i 1+ ^j ex p(" lo 8( 1+ ^l Write z = x/y/u. Then U log (l + ~J -xy/u = ^ log (1 + Z) - j log(l + *)-* -»■ — £x 2 as z -* and the result follows. 4 We use proposition 13.29 to check that the improper integral exists for x > and y > 0. The inequalities t x -\i-ty- 1 <t x - 1 (0<r<i) t x -\\-ty- 1 <(i-ty- 1 (o<r<i) suffice for this purpose. The fact that B{x, y) is a continuous function of x (and of>») is proved in the same manner as exercise 17.4(5). The fact that its logarithm is convex is proved in the same manner as exer- cise 17.4(6). 5 After the previous question, we need only show that /(l) = 1 and f(x + l) = xf(x). Now m = T ^B(l,y ) = yj;\l-t) y -' d t = l. ro) Also,ifO<a<0<l -0 -r*-(l-f) 3 y + f* xf- x -{\-tf dt. a Ja y j a m-ty-'dt = It follows that B(x + l,y) = xy~ l B(x,y + 1). But B(x,y + l) = r*V- , (l--/f<& = r^-'il-ty-^l-Odt J — *■ J - * = fi(*^)-5(x +!,>-). 250 Solutions to exercises (1 7. 7) Hence B(x+l,y) = -{B(x,y)-B(x+l,y)} 1 +- B(x + \,y) = -d(x,y) y y B(x+l,y) = x + y B(x,y). It follows that *+»-&£g&** + U* = (* +y) r(y) r(x +y) x B(x,y) = xftx). T(y) x+y Since /satisfies all the conditions of theorem 17.6, it follows that f(x) = r(x) ix > 0) and therefore Vt " Tix+y) 6 Take x = y = § in the previous question. Now O/l jU - r & r & _ p/K2 »Q,0- r(1) -I ( 2 ). On the other hand B &b-Loy/ua-t)\-L*o dt Thus r(|) = \A- Write t = x 2 ll. Then 1/2 2 sin cos 6 d6 sin flV(l - sin 2 0) = ff. Jo Hence V2 dx = "(a/r^Vd* = --r/*» = Jo ( 2? )" V2 r ® = V2 £ = 2../- - V(2ff). SUGGESTED FURTHER READING /. Further analysis J. C. Burkill and H. Burkill, ^l Second Course in Mathematical Analysis (Cambridge University Press, 1970). This book is notable in the clarity of its exposition but is perhaps a little old- fashioned in its treatment of some topics. W. Rudin, Principles of Mathematical Analysis (McGraw-Hill, 1964). This is a well-tried and popular text. The treatment is rather condensed. T. M. Aposto], Mathematical Analysis (Addison-Wesley, 1957). This is a useful text, especially in its treatment of vector methods. R. V. Churchill, Complex Variables and Applications (McGraw-Hill, 1960). All the books above include a discussion of complex analysis (i.e. analysis involving the entity i = V — 0. b ut this book is a particularly easy introduction to the topic. H. L. Royden, Real Analysis (Macmillan, 1968). This is an excellent book. The choice of material and manner of presentation has been carefully considered. Its treatment of the Lebesque integral is particularly good. It is, however, a book which makes considerable demands on the reader. G. F. Simmons, Introduction to Topology and Modern Analysis (McGraw-Hill, 1963). The title of the book describes its content. It is very well-written indeed. 2. Logic and set theory E. A. Maxwell, Fallacies in Mathematics (Cambridge University Press, 1963). This is for entertainment purposes. But it also contains some very instructive examples. N. Ya Vilenkin, Stories about Sets (Academic Press, 1968). This is another entertaining book in which the paradoxes of the infinite are explored at length. It is highly recommended. P. R. Halmos, Naive Set Theory (Van Nostrand, 1960). This is the standard reference for elementary set theory. 251 252 Suggested further reading K. Kuratowski, Introduction to Set Theory and Topology (Pergamon Press, 1972). This is a very good book which covers some difficult material. A. Margaris, First Order Mathematical Logic (Blaisdell, 1967). Formal mathematical logic is a fascinating and expanding field. This book is a good introduction. 3. Foundations of analysis E. Moise, Elementary Geometry from an Advanced Standpoint (Addison-Wesley, 1963). The real number system was developed largely in response to geometric needs. This fascinating and clearly written book describes this process and many other topics. E. Landau, Foundations of Analysis (Chelsea, 1951). This remains an excellent and simple account of the algebraic foundations of the real number system. NOTATION e.* 1 f(x) -* I as x ■* | - 74 1 /(x) ■* / as x -» | 75 c 2 /(x) -»4-°°asx- > -£ + 82 N.Z.Q.R 2 /(x) -» / as = ■* — oo 82 oo 3 fai WB 91 >,>,<,< 3 0(h) 92 M 10 /"(?). D 2 m) 92 d(x,y) 11 8y, 5x 93 sup, inf 15,70 ± 93 max, min 15 dx (fi,b),(a,b),[a.b) df 93 [a,b], (a, °°), [a, °°) dy, dx 94 (-«►,&),(-«",*] 1 /«+),/(?-) 108 d(l S) 19 S(P) no- x„ -► / as w -*• °° 28 f"f(x)dx lim x n n -» oo 28 no x n "* + °° as n "* °° 38 | f(x)dx,\ JO J-*a f(x)dx 132 x n •* — °° as n •* °° 39 lim sup x n 48 logx, Inx 137 n-»M e 137 ,141 liminfx n 48 expx 140 oa a 1 ', e x 141 E« n 53 oo n = l I a n (*-£)" 143 f.A*B 64 n=0 fix) 64 sin x, cos x 151 f(S) 65 7T 153 f + g.W.fg.f/g 68 tanx 154 f°g 68 arcsin x, arccos x, arctan x 154 r 68 a n ~ b n 156 f(x) ->lasx->Z + 74 T(x) 157 253 INDEX Abel's lemma, 170 Abel's theorem, 16 absolute convergence of series, 60 of power series, 144 absolute value, 10 analytic function, 145 anti-derivative, 124 arccosine function, 154 Archimedian property, 20 arcsine function, 154 arctangent function, 154 area, 13,119 arithmetic mean, 8 backwards induction, 23, 25 beta function, 160 binomial theorem, 25, 106, 149 bounded above, 13 below, 13 function, 70 set, 13 bounds, 13 Brouwer's fixed point theorem, 89 Cauchy mean value theorem, 104 Cauchy-Schwarz inequality, 7, 13 Cauchy sequence, 50 closed interval, 17, 19 cluster point, 52 combination theorem for functions, 79 for sequences, 30, 161 compact interval, 17, 52 comparison test, 58 composite function, 68 composition, 68 concave functions, 1 14 conditional convergence, 60 continuity, 84 et seq. at a point, 77 on an interval, 84 continuity property, 86 continuous, 77 on left, 77 on right, 77 continuum property, 141 et seq. convergence of functions, 74 of sequences, 27 of series, 53 convex functions, 1 14 cosine function, 151 decreasing function, 108 sequence, 34 degree of polynomial, 25, 67 derivative, 91, 95 differentiable, 91 differential, 93, 95 differentiation, 91 et seq. distance between points, 1 1 between point and set, 19 divergent functions, 82 sequences, 38 series, 54 domain of function, 64 element of set, 1 empty set, 1 Euler— Maclaurin formula, 134 Euler's constant, 1 39 exponential function, 140 function, 64 et seq. gamma function, 157 255 256 Index geometric mean, 23 gradient , 9 1 graph, 64 harmonic mean, 8 L'Hopital's rule, 104 image of point, 64 of set, 65 improper integral, 132 increasing function, 108 sequence, 33 induction, 22 inequalities, 3 infimum, 15,71 infinitesimal, 94 integer, 2 integral, 120 integral test, 135 integration, 1 19 el seq. by parts, 129 intermediate value theorem, 87 interval, 16 of convergence, 144 inverse function, 68, 1 10 irrational number, 2, 9 Leibniz's rule, 97 limit of function, 74 of sequence, 27; inferior, 48; superior, 48 limit point of set, 52 local maximum, 100 local minimum, 100 logarithm function, 137 lower bound, 14 maximum, 15, 71 mean value theorem, 102 minimum, 15, 71 Minkowski's inequality, 8,10 modulus, 10 monotone function, 108 sequence, 34 natural logarithm, 138 natural number, 2, 20 Newton-Raphson process, 117 nth root test, 59, 144, 164 open interval, 17, 19 oscillating sequence, 39 partial sum, 53 partition, 120 periodicity, 153 point of accumulation, 52 polynomial, 25, 66 powers, 141 power series, 143 et seq. primitive, 124 principle of induction, 22 quadratic equation, 6 radius of convergence, 144 range of function, 65 of sequence, 27 rational function, 67 rational number, 2 ratio test, 59, 144, 164 real number, 2 Riemann integral, 128 Rolle's theorem, 102 roots, 6, 112 sandwich theorem for functions, 80 for sequences, 3 1 sequence, 27 series, 53 set, 1 sine function, 151 slope, 91 stationary point, 101 Stirling's formula, 156, 160 strictly increasing or decreasing function, 108 sequence, 34 subset, 1 sum of series, 53 supremum, 15, 70 tail of series, 58 tangent function, 154 Taylor polynomial, 97 Taylor series, 145 Taylor's theorem, 105 tends to a limit, 27, 74 term of sequence, 27 transcendental number, 10 1 Index triangle inequality, 10 trigonometric function, 1 50 upper bound, 1 3 257 uniform convergence, 147 unbounded set, 14 function, 71 Jacket design by Ken Vail Printed in Great Britain Surfaces H.B.GRIFFITHS Professor Griffiths relates and explains toplogy in terms of everyday experience. Beginning with the question, 'What do we mean by surfaces?' he goes on to build up a set of definitions based on an examination of three-dimensional models and drawings, and only then to relate these to the more advanced formal mathematics. He thus provides the background for readers to inject spatial concepts into their mathematical thinking. The text is well illustrated with toned drawing which create a three-dimensional effect where appropriate and there are abundant exercises with hints to their solution. Partial Differential Equations E.T.COPSON 'Copson aims at both mathematicians and users of partial differential equations. There is, again, nothing available like this amalgam of pure theory and methods. It suddenly becomes clear as one reads how the various tricks fall into place as part of a surprisingly elegant whole ... The book will form a perfect advanced undergraduate or early postgraduate course in mathematics or physics . . . There is a good selection of examples in each chapter.' The Times Higher Education Supplement Cambridge University Press 521 21480 7